Class 12 Physics

Electromagnetic Waves (Emw) Multiple Choice Questions And Answers

Question 1. In a certain region electric field \(\vec{E}\) and magnetic field \(\vec{B}\) are perpendicular to each other. An electron enters the region perpendicular to the direction of both E and B and moves undeflected. The speed of the electron is _________.

1. \({\vec{E}} \cdot {\vec{B}}\)
2. \(|{\vec{E}} \times {\vec{B}}|\)
3. \(\frac{|{\vec{E}}|}{|{\vec{B}}|}\)
4. \(\frac{|{\vec{B}}|}{|{\vec{E}}|}\)

Answer: 1. \({\vec{E}} \cdot {\vec{B}}\)

Question 2. Which wave travels with the speed of light _________

1. Sound wave
2. Heatwave
3. Shock wave
4. Microwave

Answer: 4. Microwave

Question 3. A television network uses _________

1. Microwaves
2. High-Frequency Radio waves
3. Light waves
4. Sound waves

Answer: 2. High-frequency radio waves

Question 4. Light waves are _______

1. Longitudinal
2. Transverse
3. Both Longitudinal and transverse
4. Mechanical

Answer: 4. Mechanical

Question 5. An electromagnetic wave passing through the space is given by equations E = E₀ sin (ωt – kx), B = B₀ sin (ωt- kx) which of the following is true?

1. E₀B₀ = ωk
2. E_0ω = B₀k
3. E₀k = B_0ω
4. E_0ωk = B₀

Answer: 3. E_0ω = B₀k

Question 6. The wavelength range of Heal waves is __________

1. 400 nm to 1 nm
2. 1 mm to 700 nm
3. 0. 1 m to 1 mm
4. 700 nm to 400 nm

Answer: 2. 1 mm to 700 nm

Question 7. The maximum value of B in an electromagnetic wave is equal to 6 x 10^-8T. Thus the maximum value of \(\vec{E}\) is __________.

1. 2 Vm^-1
2. 18 Vm^-1
3. 2.5 Vm^-1
4. 6 Vm^-1

Answer: 2. 18 Vm^-1

E = cB

E = 3 x 10⁸ x 6 x 10^-8

E = 18 Vm^-1

Question 8. Two oppositely charged particles oscillate about their mean equilibrium position in free space, with a frequency of 10⁹ Hz. The wavelength of the corresponding electromagnetic wave produced is _______.

1. 0.3 m
2. 3 x 10¹⁷ m
3. 10⁹ m
4. 3.3 m

Answer: 1. 0.3 m

⇒ \(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{10^9}\)

∴ 0.3 m

Question 9. For a radiation of 6 GHz passing through air, the wave number (number of waves) per 1 m length is _______ (1 GHz = 10⁹ Hz)

1. 5
2. 3
3. 20
4. 30

Answer: 3. 20

Wave No. \(\bar{v}=\frac{1}{\lambda}\) ∵ \(\left[v=\frac{c}{\lambda}\right]\) \(\left[\frac{c}{v}=\frac{1}{\lambda}\right]\)

∴ \(\bar{v}=\frac{v}{c}=\frac{6 \times 10^9}{3 \times 10^8}=20\)

Question 10. Which one of the following is an equation of magnetic energy density?

1. \(\frac{B^2}{2 \mu_0}\)
2. \(\frac{1}{2} \mu_0 B^2\)
3. \(\frac{2 B^2}{\mu_0}\)
4. \(\frac{\mathrm{B}^2}{\mu_0}\)

Answer: 1. \(\frac{B^2}{2 \mu_0}\)

Question 11. Dimension of \(\) is same as dimension of ( where μ = magnetic constant, ε = Dielectric constant)

1. Velocity
2. Square of velocity
3. Acceleration
4. Momentum

Answer: 2. Square of velocity

Electromagnetic Waves Assertion And Reason

For questions numbers 1 to 10 two statements are given labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (1), (2), (3), and (4) as given below.

1. Both (A) and (R) are correct, (R) is the correct explanation of (A).
2. Both (A) and (R) are correct, (R) is not the correct explanation of (A).
3. (A) is correct; (R) is incorrect.
4. (A) is incorrect; (R) is incorrect

Question 1. Assertion: Electromagnetic waves do not require a medium for their propagation.

Reason: They cannot travel in a medium.

Answer: 3. (A) is correct; (R) is incorrect.

Question 2. Assertion: A changing electric field produces a magnetic field.

Reason: A changing magnetic field produces an electric field.

Answer: 2. Both (A) and (R) are correct, (R) is not the correct explanation of (A).

Question 3. Assertion: X-rays travel with the speed of light.

Reason: X-rays are electromagnetic rays.

Answer: 1. Both (A) and (R) are correct, (R) is the correct explanation of (A).

Question 4. Assertion: Environmental damage has increased the amount of ozone in the Atmosphere.

Reason: The increase of ozone increases the amount of ultraviolet radiation on the earth.

Answer: 4. (A) is incorrect; (R) is incorrect

Question 5. Assertion: Electromagnetic Radiation exerts pressure.

Reason: Electromagnetic waves carry both momentum and energy.

Answer: 1. Both (A) and (R) are correct, (R) is the correct explanation of (A).

Question 6. Assertion: The EM waves of shorter wavelengths can travel longer distances than those of longer wavelengths.

Reason: The shorter the wavelength, the larger the velocity of propagation.

Answer: 3. (A) is correct; (R) is incorrect.

Question 7. Assertion: EM waves follow the Superposition principle.

Reason: Differential expression of EM wave is linear.

Answer: 1. Both (A) and (R) are correct, (R) is the correct explanation of (A).

Question 8. Assertion: Sound waves cannot travel in a vacuum, but light waves can.

Reason: Light is an electromagnetic wave – but sound is a mechanical wave.

Answer: 1. Both (A) and (R) are correct, (R) is the correct explanation of (A).

Question 9. Assertion: Microwaves are better carriers of signals than radio waves.

Reason: Electromagnetic waves do not require any medium to propagate.

Answer: 2. Both (A) and (R) are correct, (R) is not the correct explanation of (A).

Question 10. Assertion: Transverse waves are not produced in liquids and gases.

Reason: The shorter the wavelength, the larger the velocity of propagation in air.

Answer: 3. (A) is correct; (R) is incorrect.

Electromagnetic Waves Short Questions And Answers

Question 1. The electric field of an electromagnetic wave is represented as E_x = E₀ sin (ωt + kz).

1. In which direction is the wave propagating?
2. In which direction does the magnetic field oscillate?

Answer:

1. Negative z direction
2. y direction

Question 2.

1. In which situation is there a displacement current but no conduction current?
2. Why are Microwaves considered suitable for radar systems used in aircraft navigation?

Answer:

1. In between the plates of a capacitor, during charging and discharging of a capacitor.
2. Microwaves have energy more than radio waves, so these can travel up to greater distances.

Question 3. Match the column

Answer: 1-C, 2-D, 3-B, 4-A, 5-E

Question 4.

1. Suppose that the earth’s atmosphere is absent, will the average temperature on the earth’s surface
   be higher or lower than what it is at present?
2. What is an electromagnetic constant?

Answer:

1. The average temperature will be lower due to the absence of the greenhouse effect.
2. All types of electromagnetic waves move with the same speed c = 3 x 10⁸ m/s in air or vacuum, so ‘c’ is called the electromagnetic constant.

Question 5.

1. The charging current for a capacitor is 0.25 A. What is the displacement current across its plates?
2. How are infrared waves produced? Write their one important use.

Answer:

1. 0.25 A
2. Infrared waves are produced by hot bodies due to molecular vibrations. These are used to
   treat muscular strain.

Question 6.

1. Which part of the electromagnetic spectrum is used for eye surgery?
2. Which part of the electromagnetic spectrum is blocked by protective welding glass?

Answer:

1. Ultra-violet rays arc used in Lasik Laser, for eye surgery.
2. Ultra-violet.

Question 7. Slate two properties of electromagnetic waves.

Answer:

1. All EM waves travel with the same speed c = 3 x 10⁸ m/s in air or vacuum.
2. EM waves have energy and momentum and these apply radiation pressure, on the surface on which they are made to fall.

Question 8.

1. The thin Ozone layer on top of the stratosphere is crucial for human survival. Why?
2. How can we show that em waves carry momentum?

Answer:

Question 9.

1. Which component of the electromagnetic wave is responsible for producing an optical effect?
2. Light can travel in a vacuum whereas sound cannot do so. Why?

Answer:

1. The Electric vector of the em wave is responsible.
2. Light is electromagnetic while sound is a mechanical wave.

Question 10.

1. For which wavelength our eyes are most sensitive?
2. Which of the electromagnetic waves is capable of penetrating layers of dust?

Answer:

1. 555nm i.e. yellow colour.
2. Infra-red.

Question 11. Electromagnetic waves of wavelengths λ₁,λ_2, and λ₃ are used in radar systems, water purifiers, and in remote switches of TV. respectively.

1. Identify the electromagnetic waves, and
2. Write one source for each of them.

Answer:

1. In radar systems ⇒ microwaves

In water purifies ⇒ UVrays

In remote switches in TV ⇒ Infrared rays

2. Microwave arc produced by special vacuum lubes (Klystrons, Magnetrons, and Gunn diodes)

• UV radiation is produced in welding arc and the sun is an important source of ultraviolet light.
• Infrared waves are produced by hot bodies and molecules.

Electromagnetic Waves Long Questions And Answers

Question 1. Gamma rays are used in radiotherapy to treat cancer. They are used to spot tumors. They kill the living cells and damage malignant tumors.

(1). What is the source of gamma rays?

1. Radioactive decay of the nucleus
2. Accelerated motion of charges in conducting wire
3. Hot bodies and molecule
4. Klystron valve

Answer: 1. Radioactive decay of the nucleus

(2). How is the wavelength of gamma rays

1. Low
2. High
3. Infinite
4. Zero

Answer: 1. Low

(3). Choose the one with the correct penetrating power order of radiation.

1. Alpha > beta > gamma
2. Beta > alpha > gamma
3. Gamma > beta > alpha
4. Gamma > alpha > beta

Answer: 3. Gamma > beta > alpha

(4). What is the other use of gamma rays?

1. Used to change white topaz to blue topaz
2. Used in aircraft navigation
3. Used to kill microbes
4. Checking fractures of bone

Answer: 1. Used to change white topaz to blue topaz

Question 2. X-rays are a form of electromagnetic radiation, similar to visible light. Unlike light, however, x-rays have higher energy and can pass through most of the objects, including the body. Medical x-rays are used to generate images of tissues and structures inside the body

(1). What is the most common method of preparation of X-rays?

1. Magnetron valve
2. Vibration of atoms and molecules
3. Bombardment of metal by high-energy electrons
4. Radioactive decay of the nucleus

Answer: 3. Bombardment of metal by high-energy electrons

(2). Which of the following sets of instruments or equipment can detect X-rays

1. Photocells, photographic film
2. Thermopiles, bolometer
3. Photographic film. Geiger tube
4. Geiger lube, the human eye

Answer: 3. Photographic film. Geiger tube

(3). Where do X-rays fall on the electromagnetic spectrum?

1. Between the UV region and infrared region
2. Between gamma rays and UV region
3. Between infrared and microwaves
4. Between microwaves and radio waves

Answer: 2. Between gamma rays and UV region

(4). What is the use of rays lying beyond the X-ray region in the electromagnetic spectrum

1. Used to kill microbes
2. Used to detect heat loss in insulated systems
3. Used in standard broadcast radio and television
4. Used in oncology, to kill cancerous cells.

Answer: 4. Used in oncology, to kill cancerous cells.

Alternating Current Multiple Choice Questions And Answers

Question 1. A capacitor and an inductor are connected in two different AC circuits with a bulb glowing in each circuit. The bulb glows more brightly when:

1. The number of turns in the inductor is increased
2. The separation between the plates of the capacitor is increased
3. An iron rod is introduced into the inductor
4. A dielectric is introduced into the gap between the plates of the capacitor

Answer: 4. A dielectric is introduced into the gap between the plates of the capacitor

Question 2. A pure inductor of 318mH and a pure resistor of 75 Ω arc connected in series to an AC source of 50 Hz. The voltage across the 75 Ω resistor is found to be 150V. The source voltage is:

1. 150 V
2. 175 V
3. 220 V
4. 250 V

Answer: 4. 250 V

⇒ \(\cos \phi=\frac{R}{Z}=\frac{V_R}{V} \text { or } V=V_R \times \frac{Z}{R}\)

∴ \(\frac{150}{75} \sqrt{75^2+\left(3.14 \times 318 \times 10^{-3}\right)^2}=2 \sqrt{5625 \times 9970}=2 \times 124.8 \simeq 2.50 \mathrm{~V}\)

Question 3. In an AC circuit, the applied voltage and resultant current are E = E₀ sin ωt and I = I₀ sin (ωt + π/2) respectively. The average power consumed in the circuit is:

1. B₀I₀
2. \(\frac{E_0 I_0}{2}\)
3. \(\frac{E_0 I_0}{\sqrt{2}}\)
4. Zero

Answer: 4. Zero

∴ \(P_{avg}=E_{rms} I_{rms} \cos 90^{\circ}=0\)

Question 4. In a series LCR circuit, at resonance, the current is equal to_____

1. \(\frac{\mathrm{V}}{\mathrm{R}}\)
2. \(\frac{\mathrm{V}}{\mathrm{x}_{\mathrm{c}}}\)
3. \(\frac{V}{X_{L}-X_C}\)
4. \(\frac{V}{\sqrt{R^2+\left(X_L+X_C\right)^2}}\)

Answer: 1. \(\frac{\mathrm{V}}{\mathrm{R}}\)

Question 5. The frequency of an AC source for which a 10 μF capacitor has a reactance of 1000 ohm is___

1. \(\frac{1000}{\pi} \mathrm{Hz}\)
2. 50 Hz
3. \(\frac{50}{\pi} \mathrm{Hz}\)
4. \(\frac{100}{\pi} \mathrm{Hz}\)

Answer: 3. \(\frac{50}{\pi} \mathrm{Hz}\)

⇒ \(X_C=\frac{1}{2 \pi \mathrm{fC}}=1000\)

∴ \(f=\frac{50}{\pi} \mathrm{Hz} \)

Question 6. Which one of the following statements is true:

1. An inductor has infinite resistance in a DC circuit.
2. An inductor and a capacitor cannot conduct in a DC circuit
3. A capacitor can conduct in a DC circuit but not an inductor.
4. An inductor can conduct in a DC circuit but not a capacitor.

Answer: 4. An inductor can conduct in a DC circuit but not a capacitor.

Question 7. In an A.C. circuit in 1 second current reduces to zero value 120 times. Hence the frequency of A.c current is ___________ Hz.

1. 50
2. 100
3. 60
4. 120

Answer: 3. 60

Question 8. What is the r.m.s. value of the current for A.C. current I = 100 cos (200 t + 45°)A.

1. 50√2 A
2. 100 A
3. 100√2 A
4. Zero

Answer: 1. 50√2 A

∴ \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{100}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=50 \sqrt{2} \mathrm{~A}\)

Question 9. In an R-C circuit when the charge on the plates of the capacitor is increasing, the energy obtained from the sources is stored in ___________.

1. Electric field
2. Magnetic field
3. Gravitational field
4. Both Magnetic field and gravitational field

Answer: 1. Electric field

Question 10. The output power in a step-up transformer is ___________

1. Greater than the input power
2. Equal to the input power
3. Maintained even during the power cut
4. Less than the input power

Answer: 4. Less than the input power

Question 11. The power factor for scries L-R A.C. circuit is ________.

1. \(\frac{\mathrm{R}}{\mathrm{X}_{L}}\)
2. \(\frac{X_L}{R}\)
3. \(\frac{R}{\sqrt{R^2+X_L^2}}\)
4. \(\frac{\sqrt{R^2+X_L^2}}{R}\)

Answer: 3. \(\frac{R}{\sqrt{R^2+X_L^2}}\)

Question 12. An alternating voltage given as V = 200 √2 sin 100 l (V) is applied to a capacitor of 5μF. The current reading of the ammeter will be equal to _________mA.

1. 80
2. 20
3. 40
4. 100

Answer: 4. 100

⇒ \(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_c}=\frac{V_0 / \sqrt{2}}{1 /\mathrm {\omega}{c}}=20\mathrm{\omega}{c}\)

∴  200 z 100 x 5 x 10^-6 = 10^-1 A = 100 x 10^-3 A = 100 mA

Question 13. The current of \(\frac{50}{\pi}\) frequency is passing through an A.C. circuit having a series combination of resistance R = 100 Ω and inductor L = 1H, then phase difference between voltage and current is __________.

1. 60°
2. 45°
3. 30°
4. 90°

Answer: 2. 45°

⇒ \(\phi=\tan ^{-1}\left(\frac{x_1}{R}\right)=\tan ^{-1}\left(\frac{2 \pi f L}{R}\right)\)

∴ \(\phi=\tan ^{-1}\left(\frac{2 \times \pi \times\frac{50}{\pi} \times 1}{100}\right)=45^{\circ}\)

Question 14. A coil of inductance L and resistance R is connected to an A.C. source of V volt. If the angular frequency of the A.C. source is equal to co rad s^-1, then the current in the circuit will be __________.

1. \(\frac{\mathrm{V}}{\mathrm{R}}\)
2. \(\frac{\mathrm{V}}{\mathrm{L}}\)
3. \(\frac{V}{R+L}\)
4. \(\frac{V}{\sqrt{R^2+\omega^2 L^2}}\)

Answer: 4. \(\frac{V}{\sqrt{R^2+\omega^2 L^2}}\)

Question 15. In an A.C. circuit current is 2A and voltage is 220 V and power is 44 W power factor is _________.

1. 0.10
2. 0.09
3. 1.80
4. 0.18

Answer: 1. 0.10

P_virtual = VI = 220 x 2 = 440 Watt

P_avg = P_virtual cos Φ

44 = 440 x cos Φ

∴ \(\cos \phi=\frac{1}{10}=0.1\)

Question 16. A 15 μF capacitor is connected to a 220, 50 Hz a.c. source. The value of capacitive reactance is ___________.

1. 424
2. 106
3. 212
4. 21.2

Answer: 3. 212

⇒ \(X_C=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 1.5 \times 10^{-6}}\)

∴ X_c = 212.314

Question 17. A power transmission line feeds input power at 3300 V to a step-down transformer with its primary windings having 2000 turns. What should be the number of turns in the secondary to gel output power at 330 V?

1. 200
2. 400
3. 33
4. 40

Answer: 1. 200

⇒ \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\)

⇒ \(\frac{330}{3300}=\frac{N_s}{2000}\)

⇒ \(\frac{1}{10}=\frac{N_s}{2000}\)

∴ N_s = 200

Alternating Current Assertion and Reason

For question numbers 1 to 7 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1). (2). (3) and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: In a Series LCR circuit connected to an AC source, resonance can take place.

Reason: At resonance XL = XC

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

Question 2. Assertion: A transformer is used to increase or decrease AC voltage only.

Reason: A transformer works based on mutual Induction.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 3. Assertion: Average power loss in scries LC circuit is always zero.

Reason: The average value of voltage and current in A.C. is zero.

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

Question 4. Assertion: The capacitor serves as a block for D.C, and offers an easy path to AC.

Reason: Capacitive reactance is inversely proportional to frequency.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 5. Assertion: When capacitive reactance is smaller than the inductive reactance in the scries LCR
circuit, voltage leads the current.

Reason: In a series LCR circuit inductive reactance is always greater than capacitive reactance.

Answer: 3. A is true but R is false

Question 6. Assertion: In the series LCR circuit, the impedance is minimal at resonance.

Reason: The currents in the inductor and capacitor arc same in the scries LCR circuit.

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

Question 7. Assertion: In series LCR circuit phase difference between current and voltage is never zero.

Reason: Voltage and current are never in phase.

Answer: 4. A is false and R is also false

Alternating Current Short Questions And Answers

Question 1. An alternating current I = (10 A) sin (100 πt) is passed through a resistor of 20 Ω. What is the average power consumed by the resistor over a complete cycle?

Answer:

P_avg= V_rms I_rms cos = (I_rms R) I_rms [cosΦ= 1]

So, P_avg = I²_rms R-C

⇒ \(\frac{I_{0}^2}{2} \times R\)

⇒ \(\frac{10 \times 10 \times 20}{2}\)

P_avg = 1000 watt

Question 2. Define ‘quality factor’ at resonance in series LCR circuit. What is its SI unit?

Answer:

The Q factor of the series resonant circuit is defined as the ratio of the voltage developed across the inductor or capacitor at resonance to the applied voltage, which is the voltage across R.

⇒ \(Q=\frac{I X_1}{I R}=\frac{{\omega}_0 L}{R}=\frac{{\omega}_1}{{\omega}_2-{\omega}_1}\)

It is dimensionless, hence it has no units. It represents the sharpness of resonance.

Question 3. An a.c. source of voltage V = V₀sinωt is connected to an ideal inductor. Draw graphs of voltage V and current 1 versus cot.

Answer:

Question 4. Explain why current Hows through an ideal capacitor when it is connected to an a.c. source but not when it is connected to a D.C. source in a steady state.

Answer:

When AC is connected to the capacitor, due to continuous change of polarity of the applied voltage there will be continuous change of polarity of capacitor plates. This causes the charge to flow across the capcitor.

In steady state, the capacitor acts as an open circuit as reactance offered by it to flow of dc (f = 0) is infinite, As \(X_c=\frac{1}{2 \pi f C}=\infty\)

Question 5.

An LCR series circuit is connected to an AC source. If the angular resonant frequency of the circuit is coo, will the current lead or lag or be in phase with the voltage when to ω < ω₀  and why?

1. We cannot step up the DC voltage using a transformer. Why?
2. On what principle does a metal detector work?

Answer:

1. at ω < ω₀

X_L< X_C

so current leads the voltage

2. For d.c f = 0

So. there is no mutual induction and the transformer works on the principle of mutual induction.

3. The metal detector works on the principle of resonance in AC circuits.

Question 6.

1. In an LCR series circuit connected to an AC source, the voltage and the current are in the same phase. If the capacitor is filled with a dielectric, will the current lead or lag or remain in phase with the voltage? Explain.
2. In the circuit, why is the rms value of net voltage not equal to the sum of voltage drops
   across individual elements?
3. Draw a graph showing variation of the impedance of the circuit with the frequency of the
   applied voltage.

Answer:

1. Given in question capacitor is filled with a dielectric slab, So new capacity increases as

⇒ \(C=\frac{\varepsilon_1 \varepsilon_0 A}{d}\)

So, C increase then X_C decreases

So, X_L > X_C

Current lag voltage, as the circuit is inductive.

2. Voltage across R, L, and C are at different phase angles. So. we must do vector addition of voltages then we get Net voltage.

3.

Question 7. A capacitor of unknown capacitance, a resistor of 100 Ω, and an inductor of self-inductance l, = (4/π²) henry are connected in series to an AC source of 200 V and 50 Hz. Calculate the value of the capacitance and impedance of the circuit when the current is in phase with the voltage. Calculate the power dissipated in the circuit.

Answer:

Current in phase with voltage means the angle between cmf and current is zero. This is a resonance condition. So

Inductive reactance = Capacitive Reactance

⇒ \(X_L=X_C \Rightarrow \omega L=\frac{1}{\omega C}\)

⇒ \(\mathrm{C}=\frac{1}{{\omega}^2 \mathrm{~L}} \text {, given } \mathrm{L}=4 / \pi^2 \quad f=50 \mathrm{~Hz}, \mathrm{~V}=200 \mathrm{~V}\)

So \(C=\frac{1 \times \pi^2}{(2 \pi)^2 \times 4 \times(50)^2} \Rightarrow C=25 \mu \mathrm{F}\)

Power dissipated \(P=V^2 / R=\frac{200 \times 200}{100}\)

∴ P = 400 W

Question 8. A series LCR circuit connected to a variable frequency 230 V source.

1. Determine the source frequency which drives the circuit in resonance.
2. Calculate the impedance of the circuit and amplitude of current at resonance.
3. Show that potential drop across LC combination is zero at resonating frequency.

Answer:

1. At Resonance

⇒ \(X_L=X_C \Rightarrow \omega_{\mathrm{r}} L=\frac{1}{\omega_{\mathrm{r}} \mathrm{C}}\)

⇒ \(f_r=\frac{1}{2 \pi \sqrt{L C}} \Rightarrow f_r=\frac{1}{2 \pi \sqrt{5 \times 80 \times 10^{-6}}}\)

⇒ \(f_r=\frac{1}{2 \pi \times 20 \times 10^{-3}}\)

∴ \(\mathrm{f}_{\mathrm{r}}=\frac{25}{\pi} \mathrm{Hz}\)

2. Impedance of the circuit at resonance

Z = R = Z = 40 Ω

Amplitude of current at resonance

⇒ \(V_{\mathrm{rms}}=I_{\mathrm{rm}} \mathrm{Z} \Rightarrow I_{\mathrm{rms}}=\frac{230}{40} \mathrm{A}\)

∴ \(I_{\mathrm{rms}}=\frac{23}{4} \mathrm{~A}\)

Amplitude, I₀ = √2 I_rms

∴ I₀ = 8.1 A

3. Potential drop across LC combination

V_LC = V_L-V_C

= I(X_L-X_C)

At resonance X_L = X_C => X_L– X_C = 0

V_C= 0

Question 9.

1. When an AC source is connected to an ideal capacitor, show that the average power
   supplied by the source over a complete cycle is zero.
2. A bulb is connected in series with a variable capacitor and an A.C. source as shown. What happens to the brightness of the bulb when the key is plugged in and the capacitance of the capacitor is gradually reduced?

Answer:

1. Given V = V₀ sin t

q = CV

q = CV₀ sin t

⇒ \(\frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{CV}_0 \quad \frac{\mathrm{d}}{\mathrm{dt}} \sin \omega t\)

I = CV₀ (cos ω t)

⇒ [\atex]\left.I=\frac{V_0}{\left(\begin{array}{c}\frac
{1}{\omega c}
\end{array}\right)} \cos {\omega} t \Rightarrow I=\frac{V_0}{X_c} \sin ({\omega} t+\pi / 2\right)[/latex]

Here, \(X_C=\frac{1}{{\omega} C} \text { and } \mathrm{I}_0=\frac{V_0}{X_C}\)

Average power

⇒ \(P_{a v}=\int_0^1 V I d t=\frac{V_0^2}{X_C} \int_0^1(\sin {\omega} t)(\sin {\omega} t+\pi / 2) d t=\frac{V_0^2}{X_c} \int_0^1(\sin \omega t)(\cos {\omega}t) d t\)

⇒ \(P_{\mathrm{av}}=\frac{\mathrm{V}_0^2}{2 X_C} \int_0^{\mathrm{T}} \sin (2{\omega} t) d t \quad\left\{\int_0^{\mathrm{T}} \sin (2{\omega} t) d t=0\right.\)

P_av = 0

2. When the AC source is connected, the capacitor offers capacitive reactance X_C = 1/Cω. The
current flows in the circuit and the lamp glows. On reducing C, X_C increases and currently
reduces, Therefore, the bulb’s glow reduces.

Question 10. A capacitor (C) and resistor (R) are connected in series with an AC source of voltage of frequency 50 Hz. The potential difference across C and R are 120 V and 90 V respectively, and the circuit’s current is 3 A. Calculate

1. The impedance of the circuit
2. The value of the inductance, which when connected in series with C and R will make the power factor of the circuit unity.

Answer:

Given: f = 50 Hz, I = 3A, V_C = 120 V and V_R = 90 V

1. Impedance of the circuit:

⇒ \(\mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}} \Rightarrow \mathrm{Z}=\frac{\sqrt{\mathrm{V}_{\mathrm{c}}^2+\mathrm{V}_{\mathrm{R}}^2}}{\mathrm{I}}\)

⇒ \(Z=\frac{\sqrt{(120)^2+(90)^2}}{3}\)

Z = 50

2. Power factor (cos Φ) = l. This is the condition of resonance. Let inductance (L) is connected in series with C and R. At resonance.

X_L = X_C

V_C = IX_C

120 = 3 X_C

X_C = 40

X_L = 40 Ω ⇒ ωL = 40 Ω ⇒ 2πfL = 40

π \(\mathrm{L}=\frac{40}{2 \pi f}=\frac{40}{2 \pi \times 50}=\frac{0.4}{\pi} \mathrm{H}\)

Question 11.

1. When an AC source is connected to an ideal inductor show that the average power supplied
   by the source over a complete cycle is zero.
2. A lamp is connected in series with an inductor and an AC source. What happens to the lamp’s brightness when the key is plugged in and an iron rod is inserted inside the inductor? Explain.

Answer:

1. Given

V = V₀ sinωt

⇒ \(\mathrm{V}=\mathrm{L} \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}\) (induced emf)

⇒ \(\mathrm{dI}=\frac{\mathrm{V}}{\mathrm{L}} \mathrm{dt}\)

⇒ \(\mathrm{dI}=\frac{\mathrm{V}_0}{\mathrm{~L}} \sin \omega t \mathrm{dt}\)

By integration \(I=\frac{-V_0}{{\omega}} \cos {\omega} t\)

∴ \(I=-\frac{V_0}{\omega L} \sin \left[\frac{\pi}{2}-\omega t\right] \Rightarrow I_0 \sin \left[\omega t-\frac{\pi}{2}\right]\)

where \(I_0=\frac{V_0}{{\omega} \mathrm{L}}\)

Average power

⇒ \(P_{\mathrm{av}}=\int_0^{\mathrm{T}} V I \mathrm{dt}\)

⇒ \(-\frac{V_0^2}{\omega L} \int_0^T \sin {\omega} t \cos {\omega}t dt\)

⇒ \(-\frac{V_0^2}{2 \omega L} \int_0^{\mathrm{T}} \sin (2 \omega \mathrm{t}) \mathrm{dt} \quad\left\{\int_0^{\mathrm{T}} \sin (2 \omega t) \mathrm{d} t=0\right.\)

= 0

2. When an iron rod is inserted into the inductor, the self-inductance of the inductor increases. On increasing L. X_L increases and current reduces. Therefore glow of the bulb reduces.

Alternating Current Long Questions And Answers

Question 1. When a pure resistance R, pure inductor L, and an ideal capacitor of capacitance C are connected in series to a source of alternating e.m.f., then-current at any instant through the three elements has the same amplitude and is represented as I = I₀ sin ωt.

However, the voltage across each element has a different phase relationship with the current as shown in the graph. The effective resistance of the RLC circuit is called the impedance (Z) of the circuit and the voltage leads the current by a phase angle Φ.

A resistor of 12Ω a capacitor of reactance 14Ω and a pure inductor of inductance 0.1 H are joined in series and placed across 200 V, 50 Hz a.c. supply.

(1). What is the value of inductive reactance?

Answer:

X_L = 2πfL = 2 x 3.14 x 50 x 0.1 = 31.4 Ω

(2). What is the value of impedance?

Answer:

∴ \(Z=\sqrt{R^2+\left(X_{L}-X_C\right)^2}=\sqrt{(12)^2+(31.4-14)^2}=21.13 \Omega\)

(3). What is the value of current in the circuit?

Answer:

∴ \(I=\frac{e}{Z}=\frac{200}{21.13}=9.46 \mathrm{~A}\)

(4). What is the value of the phase angle between current and voltage?

Answer:

∴ \(\tan \phi=\frac{X_L-X_C}{R}=\frac{31.4-14}{12}=1.45 \Rightarrow \phi=\tan ^{-1}(1.45)\)

Question 2. The power averaged over one full cycle of a.c. is known as average power. It is also known as true power \(P_{\mathrm{av}}=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi=\frac{V_0 I_0}{2} \cos \phi\)

Root mean square or simply rms watts refer to continuous power.

A circuit containing an 80 mH inductor and a 60 μF capacitor in series is connected to a 230V, 50 Hz supply. The resistance of the circuit is negligible.

(1). What is the average power transferred to the inductor?

Answer:

Zero

(2). What is the total average power absorbed by the circuit?

Answer:

This is an LC circuit so the average power absorbed by the circuit is zero.

(3). Find the value of current amplitude.

Answer:

⇒ \(I=\frac{e}{Z}=\frac{c}{X_L-X_c}\)

e = 230 V0

X_L = L = 2πfL

= 2 x 3.14 x 50 x 80 x 10^-3 = 25.120 Ω

⇒ \(X_C=\frac{1}{{\omega} C}=\frac{1}{2 \pi fC}\)

∴  \(\frac{1}{2 \times 3.14 \times 50 \times 60 \times 10^{-6}}=53 \Omega\)

So, \(I=\frac{230}{53-25.120}=8.249 \mathrm{~A}\)

I₀ = √2I_rms

= 2 x 8.249 = 11.6 A

(4). Find the rms value of current.

Answer:

⇒ \(I=\frac{e}{Z}=\frac{e}{X_L-X_C}\)

e = 230 V0

x_L = L = 2πfL

= 2 x 3.14 x 50 80 x 10^-3 = 25. 120

⇒\(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)

⇒ \(\frac{1}{2 \times 3.14 \times 50 \times 60 \times 10^{-6}}=53 \Omega\)

So, \(I=\frac{230}{53-25.120}=8.249 \mathrm{~A}\)

Question 3.

1. In a series LCR circuit connected to an a.c. source of voltage V = V_m sinωt, use phasor diagram to derive an expression for the current in the circuit. Hence obtain the expression tor the power dissipated in the circuit. Show that power dissipated at resonance is maximum.
2. In a series LR circuit, X_L = R, and the power factor of the circuit is P₁. When a capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. Calculate P₁/P₂.

Answer:

1. Suppose OA, OB, and OC represent the magnitude of phasor V_R, V_L, and V_C respectively. In the case of V_L > V_C, the resultant of (V_R) and (V_L-V_C), is represented by OE. Thus from ΔOAE

⇒ \(\mathrm{OE}=\sqrt{\mathrm{OA}^2+\mathrm{AE}^2}\)

⇒ \(V=\sqrt{V_R^2+\left(V_L-V_C\right)^2}\)

Substituting the value of VR, VL, and VC we have

⇒ \(V=\sqrt{(I R)^2+\left(I X_L-I X_C\right)^2}\)

or \(I=\frac{V}{\sqrt{(R)^2+\left(X_L-X_C\right)^2}}\)

The effective opposition offered by L, C, R to a.c. supply is called the impedance of the LCR circuit and is represented by Z.

∴ \(I=\frac{V}{Z}\)

So, comparing \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

Also from ΔOAE

⇒ \(\tan \phi=\frac{A E}{O A}=\frac{V_L-V_c}{V_R}\)

or \(\tan \phi=\left(X_L-X_C\right) / R\)

or \(\phi=\tan ^{-1} \frac{\left(X_L-X_C\right)}{R}\)

Power dissipation in LCR circuit:

The instantaneous power supplied by the source is

P = VI

⇒ \(P=\left(V_m \sin \omega t\right) \times i_m \sin (\omega t+\phi) = \frac{V_m i_m}{2}[\cos \phi-\cos (2 \omega t+\phi)]\)

[2sinAsinB = cos(A-B)-cos(A+B)]

For Average power, the second term becomes zero in the complete cycle.

So \(P_{a v}=\frac{V_m i_m}{2} \cos \phi \quad \int_0^{\mathrm{T}} \cos (2 \omega t+\phi) d t=0\)

So \(P_{\mathrm{av}}=\frac{V_{\mathrm{m}}}{\sqrt{2}} \frac{i_{\mathrm{m}}}{\sqrt{2}} \cos \phi=V_{\mathrm{rms}} i_{\mathrm{rms}} \cos \phi\)

So \(\mathrm{P}_{\mathrm{av}}=\mathrm{V}_{\mathrm{rms}} \mathrm{i}_{\mathrm{rms}} \cos \phi\)

At resonance condition, cosΦ = 1 (because Φ) = 0), R becomes the effective impedance of a circuit. So power dissipated is maximum at resonance condition

2. In series LCR circuit impedance

⇒ \(Z=\sqrt{R^2+\left(X_L-X_C\right)^2}\) and power factor p = R/Zero

case 1: In LR circuit When XL = R, So Z = (2R2)1/2

Z = √2R = Now \(P_1=\frac{R}{Z}=\frac{R}{\sqrt{2} R} \Rightarrow P_1=\frac{1}{\sqrt{2}}\)

Case 2: X_L = X_C, Z=R, So power factor P₂ becomes equal to 1

P₂ = 1

So ration \(\frac{P_1}{P_2}=\frac{1}{\sqrt{2}}: \frac{1}{1} \Rightarrow \frac{P_1}{P_2}=\frac{1}{\sqrt{2}}\)

Question 4. A 2 μF capacitor, 100 Ω resistors, and 8 H inductor are connected in series with an AC source.

1. What should be the frequency of the source such that the current drawn in the circuit is maximum, What is this frequency called?
2. If the peak value of c.m.f. of the source is 200 V, find the maximum current.
3. Draw a graph showing the variation of amplitude of circuit current with changing frequency of applied voltage in a series LCR circuit for two different values of resistance R₁ and R₂ (R₁ > R₂).
4. Define the term ‘Sharpness of Resonance’. Under what conditions, does a circuit become
   more selective?

Answer:

1. Source frequency, when current is maximum is given by

⇒ \(f=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{8 \times 2 \times 10^{-6}}}\) [L = 8H and C = 2μF]

∴ \(f=\frac{1}{2 \pi \times 4 \times 10^{-3}}\)

f = 39.80 Hz

The frequency at which the current maximum, is called resonant frequency.

2. Given E₀ = 200V, R= 100Ω

∴ \(I_{\max }=\frac{E_0}{R}=\frac{200}{100}=2 \mathrm{~A}\)

3.

4. The sharpness of resonance is given by the quality factor (Q factor) of a resonant circuit. It is defined as the ratio of the voltage drop across the inductor (or capacitor) to the applied voltage. Sharper the curve, the circuit will be more selective. For resistant R₂, the circuit is more selective.

Question 5.

1. Draw a labeled diagram of a step-down transformer. State the principle and its working.
2. Express the turn ratio in terms of voltages.
3. Find the ratio of primary and secondary currents in terms of turn ratio in an ideal
   transformer.
4. How much current is drawn by the primary of a transformer connected to a 220 V supply when it delivers power to a 1 10 V- 550 W refrigerator?

Answer:

1.

Principle: The Transformer works on the principle of mutual induction, in which an EMF is induced in the secondary coil by changing the magnetic flux in the primary coil.

Working: When an alternating current source is connected to the ends of the primary coil, the current changes continuously in the primary coil, due to which magnetic flux linked with the secondary coil changes continuously. Therefore, the alternating emf of the same frequency is developed across the secondary terminals.

2. \(\frac{N_s}{N_p}=\frac{V_s}{V_p} \quad\left\{\frac{N_s}{N_p}=\right.\text { turn ratio }\)

3. For ideal transformer

Output power = Input power

V_SI_S = V_PI_P

⇒ \(\frac{V_S}{V_p}=\frac{I_p}{I_S} \quad\left\{\frac{V_s}{V_p}=\frac{N_s}{N_p}\right.\)

∴ \(\frac{N_s}{N_p}=\frac{I_p}{I_s}\)

4. Given V_P = 220 V, V_S = 1 1 0 V. P = 550 W. I_P = ?

⇒ \(I_p=\frac{\text { Power }}{\text { Primary Voltage }}=\frac{P}{V_p}\)

∴ \(\mathrm{I}_{\mathrm{P}}=\frac{550}{220}=2.5 \mathrm{~A}\)

Question 6. A device ‘X’ is connected to an AC source V = V₀ sin ωt. The variation of voltage, current, and power in one cycle is shown in the following graph:

1. Identify the device ‘X’.
2. Which of the curves A, B, and C represent the voltage, current, and power consumed in the circuit? Justify your answer.
3. How does its impedance vary with the frequency of the AC source? Show graphically.
4. Obtain an expression for the current in the circuit and its phase relation with AC voltage.

Answer:

1. Capacitor

2. Curve: A represents power because in a pure capacitive AC circuit, power consumed in one cycle is zero and the frequency of power is twice the frequency of voltage (or current). CurvcB represents voltage. Curve-C represents current because in a pure capacitive AC circuit, current leads the voltage by π/2.

3. \(Z=X_C \Rightarrow Z=\frac{1}{2 \pi f C}\)

4. We know, \(V=\frac{q}{C} \text { or } q=C V\)

Also, q = CV₀ sin t

or \(\frac{\mathrm{dq}}{\mathrm{dt}}=C \mathrm{~V}_0 \frac{\mathrm{d}}{\mathrm{dt}}(\sin {\omega} \mathrm{t})\)

I = CV₀(cos ωt)

∴ \(I=\frac{V_0}{\left(\begin{array}{c}
\frac1{\omega C}
\end{array}\right)} \cos \omega t\)

\(I=\frac{V_0}{X_c} \cos {\omega} t\) (\(x_C=\frac{1}{\omega C}\))

\(I=I_0 \sin (\omega t+\pi / 2)\)     \(\left[I_0=\frac{V_0}{X_C}\right]\)

∴ Leads by (π/2) with voltage.

Question 7. A device X is connected across an AC source of voltage V = V₀ sinωt. The current through X is given as \(I=I_0 \sin \left(\omega t +\frac{\pi}{2}\right)\)

1. Identify the device X and write the expression for its reactance.
2. Draw graphs showing the variation of voltage and current with lime over one cycle of ac, for X.
3. How does the reactance of device X vary with the frequency of the AC? Show this variation
   graphically.
4. Draw the phasor diagram for the device X.

Answer:

1. X: Capacitor

Reactance \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)

2.

3. Reactance of the capacitor varies in inverse proportion to the frequency i.e \(x_c \propto \frac{1}{f}\)

4.

Electric Charges And Fields Multiple Choice Questions And Answers

Question 1. An electric dipole placed in a non-uniform electric field will experience:

1. Only a force
2. Only a torque
3. Both force and torque
4. Neither force nor torque

Answer: 3. Both force and torque

Question 2. Let N₁ be the number of electric field lines going out of an imaginary cube of side a that encloses an isolated point charge 2q and N₂, be the corresponding number for an imaginary sphere of radius a that encloses an isolated point charge 3q Then (N₁/N₂) is :

1. 1/π
2. 2/3
3. 9/4
4. 5/3

Answer: 2. 2/3

∴ \(\phi=\frac{q}{\varepsilon_0}\)

So, \( N_1 =\frac{2 q}{\varepsilon_0} \)

∴ \(N_2 =\frac{3 q}{\varepsilon_0} \Rightarrow \frac{N_1}{N_2}=\frac{2}{3}\)

Question 3. Let F₁ be the magnitude of the force between two small spheres, charged to a constant potential In free space, and F₂, be the magnitude of the force between them in a medium of dielectric constant K. Then (F₁/F₂) is:

1. \(\frac{1}{K}\)
2. K
3. K²
4. \(\frac{1}{\mathrm{~K}^2}\)

Answer: 2. k

⇒ \( F_m=\frac{F_0}{K} \)

⇒ \( F_2=\frac{F_1}{K}\)

So, \( $\frac{F_1}{F_2}=K\)

Question 4. A charge Q is placed at the center of the line joining two charges q and The system of the three charges will be in equilibrium if Q is :

1. \(+\frac{q}{3}\)
2. \(-\frac{q}{3}\)
3. \(+\frac{q}{4}\)
4. \(-\frac{q}{4}\)

Answer: 4. \(-\frac{q}{4}\)

According to question

F_net = 0

According to q at AB

F_AC = F_AB

⇒ \(\frac{\mathrm{kqQ}}{\left(\frac{\mathrm{d}}{2}\right)^2}=\frac{\mathrm{kqq}}{\mathrm{d}^2}\)

4Q = q

⇒ \(\mathrm{Q}=\frac{\mathrm{q}}{4}\)

Q must be negative

so, \(\mathrm{Q}=\frac{\mathrm{-q}}{4}\)

Question 5. Electric flux of an electric field E through an area d\(\vec{A}\) is given by :

1. \({\vec{E}} \times \mathrm{d} \vec{A}\)
2. \(\frac{\vec{E} \times \mathrm{d} \overrightarrow{\vec{A}}}{\varepsilon_0}\)
3. \({\vec{E}} \cdot \mathrm{d}{\vec{A}}\)
4. \(\frac{\vec{E} \cdot d \vec{A}}{\varepsilon_0}\)

Answer: 3. \({\vec{E}} \cdot \mathrm{d}{\vec{A}}\)

Question 6. Two point charges +16 q and -4 q are located at x = 0, and x = L. The location of the point on the x-axis at which the resultant electric field due to these charges is zero is:

1. 8L
2. 6L
3. 4L
4. 2L

Answer: 4. 2L

Question 7. An electric dipole of dipole moment 4 x 10^-5C-m, kept in a uniform electric field of 10^-3 NC^-1 experiences a torque of 2 x 10^-8 Nm. The angle that the dipole makes with the electric field is:

1. 30°
2. 45°
3. 60°
4. 90°

Answer: 1. 30°

τ = pE sin θ

Question 8. Three identical charges arc placed at the x-axis from left to right with adjacent charges separated by a distance d. The magnitude of the force on a charge from its nearest neighbor charge is F. Let \(\hat{i}\) be the unit vector along + x-axis. then the net force on each charge from left to right is :

1. \((2 F \hat{i},-2 F \hat{i}, 2 F \hat{i})\)
2. \((\mathrm{F} \hat{\imath}, 0, \mathrm{~F} \hat{\imath})\)
3. \((-5 / 4 F \hat{\imath}, 0,+5 / 4 F \hat{\imath})\)
4. \((2 F \hat{\imath}, 0,2 F \hat{\imath})\)

Answer: 3. \((-5 / 4 F \hat{\imath}, 0,+5 / 4 F \hat{\imath})\)

Question 9. A lest charge of 1.6 x 10^-19C is moving with a velocity \({\vec{v}}=(4 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}) \mathrm{ms}^{-1}\) in a magnetic field \({\vec{B}}=(3 \hat{\mathrm{k}}+4 \hat{\mathrm{i}}) \mathrm{T}\) The force on this lest charge is:

1. \(24 \hat{j} \mathrm{~N}\)
2. \(-24 \hat{i} \mathrm{~N}\)
3. \(24 \hat{k} \mathrm{~N}\)
4. 0

Answer: 4. 0

Question 10. If a charge is moved against a coulomb force of an electric field, then the

1. The intensity of the electric field increases
2. The intensity of the electric field decreases
3. Work is done by the electric field
4. Work is done by the external source

Answer: 4. Work is done by the external source

Question 11. A charge Q is located at the center of a circle of radius r. The work done in moving a test charge q₀ from point A to point B (at opposite ends of diameter AB) to complete a semicircle is \(\left[k=\frac{1}{4 \pi \varepsilon_0}\right]\)

1. \(k \frac{q_0 Q}{r}\)
2. \(k \frac{q_0 Q}{r^2}\)
3. kq₀Qr
4. Zero

Answer: 4. Zero

Question 12. Two charged spheres A and B having their radii in ratio 1: 2 are connected with a conducting wire, the ratio of their surface charge densities (σ_A/ σ_B) will be:

1. \(\frac{1}{2}\)
2. 2
3. \(\frac{1}{4}\)
4. 4

Answer: 2. 2

Question 13. The force acting between two point charges kept at a certain distance is 5 N. Now the magnitudes of charges arc doubled and the distance between them is halved, the force acting between them is _______ N.

1. 5
2. 20
3. 40
4. 80

Answer: 4. 80

\(F=\frac{k q \cdot q}{r^2}=5 \)

Now, \(F=\frac{k(2 q)(2 q)}{\left(\begin{array}{l}
r \\
2
\end{array}\right)^2}\) = 16 x 5 = 80 N

Question 14. When an electron and a proton are placed in an electric field _________.

1. The electric forces acting on them are equal in magnitude as well as direction.
2. Only the magnitudes of forces are the same.
3. Accelerations produced in them are the same
4. The magnitudes of accelerations produced in them are the same

Answer: 2. Only the magnitudes of forces are the same.

Question 15. Two spheres carrying charges q arc hanging from, the same point of suspension with the threads of length 2 m, in space free from gravity. The distance between them will be, ______ m.

1. 0
2. 1.0
3. 4.0
4. 2.0

Answer: 3. 4.0

Question 16. When two spheres having 2Q and -Q are placed at a certain distance. The force acting between them is F. Now these are connected by a conducting wire and again separated from each other. How much, force will act between them if the separation, now is the same as before?

1. F
2. \(\frac{F}{2}\)
3. \(\frac{F}{4}\)
4. \(\frac{F}{8}\)

Answer: 4. \(\frac{F}{8}\)

⇒ \(\mathrm{F}=\frac{2 \mathrm{k} \mathrm{Q}^2}{\mathrm{r}^2}\) →(1)

These are connected after the separation

⇒ \(\mathrm{F}^{\prime}=\frac{\mathrm{k}\left(\frac{\mathrm{Q}}{2}\right)\left(\frac{\mathrm{Q}}{2}\right)}{\mathrm{r}^2}=\frac{\mathrm{kQ}^2}{4 \mathrm{r}^2}\) →(2)

Equation (2)÷(1)

∴ \(\frac{F^{\prime}}{F}=\frac{1}{8} \Rightarrow F^{\prime}=\frac{F}{8}\)

Question 17. When a 10 μC charge is enclosed by a closed surface, the flux passing through the surface is Φ. Now another -5μC charge is placed inside the same closed surface, then the flux passing through the surface is ________.

1. 2Φ
2. Φ/2
3. Φ
4. Zero

Answer: 2. Φ/2

Question 18. An electric dipole is placed in a uniform electric field. The resultant force acting on it is _________.

1. Always zero
2. Never zero
3. Depends on the relative position
4. Depends upon the dipole moment

Answer: 1. Always zero

Question 19. Electric field due to a dipole at a large distance (r) falls off as ________.

1. \(\frac{1}{r}\)
2. \(\frac{1}{r^2}\)
3. \(\frac{1}{r^3}\)
4. \(\frac{1}{r^4}\)

Answer: 3. \(\frac{1}{r^3}\)

⇒ \(E=\frac{2 k p}{r^3}\) (On axis)

∴ \(\mathrm{E}=\frac{\mathrm{kp}}{\mathrm{r}^3}\) (on equator for large distance)

Question 20. The resultant force and resultant torque acting on an electric dipole kept in a uniform electric Held (θ ≠ 0° or 180°) are \(\vec{F}\) and \(\vec{\tau}\) then:

1. \({\vec{F}} \neq 0 ; \quad \vec{\tau}=0 \)
2. \({\vec{F}}=0 ; \quad \vec{\tau} \neq 0\)
3. \({\vec{F}}=0 ; \vec{\tau}=0\)
4. \(\dot{\vec{F}} \neq 0 ; \vec{\tau} \neq 0\)

Answer: 2. \({\vec{F}}=0 ; \quad \vec{\tau} \neq 0\)

Question 21. The liquid drop of mass ‘m’ has a charge ‘q’. What should be the magnitude of electric field E to balance this drop?

1. \(\frac{E}{m}\)
2. \(\frac{\mathrm{mg}}{\mathrm{q}}\)
3. mgq
4. \(\frac{\mathrm{mq}}{\mathrm{g}}\)

Answer: 2. \(\frac{\mathrm{mg}}{\mathrm{q}}\)

∴ mg = qE

Question 22. The number of electric field lines that emerged from 1 mC charge is _____.

1. 1.13×10²
2. 9×10⁹
3. 1.13×10¹¹
4. 9×10^-9

Answer: 1. 1.13×10²

∴ \(\phi=\frac{\mathrm{q}}{\varepsilon_0}=\frac{1 \mathrm{mC}}{8.85 \times 10^{-12}}=1.13 \times 10^8 \frac{\mathrm{N}}{\mathrm{C}} \times \mathrm{m}^2\)

Question 23. A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, the outward electric flux will:

1. Increase four times
2. Be reduced to half
3. Remains The same
4. Be doubled

Answer: 3. Remains The same

Assertion And Reason

For questions numbers 1 to 4 two statements are given-one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (1), (2), (3), and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: In electrostatics, electrostatic field lines can never be closed loops.

Reason: The number of electric field lines originating from or terminating on a charge is
proportional to the magnitude of the charge.

Answer: 2. A is true but R is false

Question 2. Assertion: Under electrostatic conditions net electric field inside a solid conductor will be zero.

Reason: Under electrostatics conditions, there will be no free electrons inside a conductor.

Answer: 3. Both A and R are true but R is NOT the correct explanation of A

Question 3. Assertion: Gauss law shows diversion when inverse square law is not obeyed.

Reason: Gauss law is a consequence of conservation of charge.

Answer: 3. A is true but R is false

Question 4. Assertion: The electrostatic force between two charges is a nonconservative force.

Reason: Electric force between two charges proportional to the square of distance between the two.

Answer: 4. A is false and R is also false

Question 5. Assertion: Electrostatic field lines are perpendicular to the surface of the conductor.

Reason: The surface of a conductor is equipotential.

Answer: 1. Both A and R are true and R is the correct explanation of A

Short Questions And Answers

Question 1. Derive an expression for the work done in rotating a dipole from the angle θ₀ to θ₁, in a uniform electric field.

Answer:

As we know, when a dipole is placed in a uniform electric field, the net force on the dipole is zero but it experiences a torque, which can be given as, \(\vec{\tau}=\vec{p} \times \vec{E}\)

This torque rotates the dipole unless it is placed parallel or anti-parallel to the external field. If we apply an external and opposite torque, it neutralizes the effect of this torque given by τ_ext and it rotates the dipole from the angle θ₀ to an angle θ₁ at an infinitesimal angular speed without any angular acceleration.

The amount of work done by the external torque can be given by

∴ \(\mathrm{W}=\int_{0_{0}}^{0_1} \tau_{\mathrm{cxt}} \mathrm{d} 0=\int_{0_0}^{0_1} p E \sin 0 \mathrm{~d} 0=p \mathrm{E}\left(\cos \theta_0-\cos \theta_1\right)\)

Question 2.

1. Draw the pattern of electric field lines due to an electric dipole.
2. Write any two properties of electric field lines.

Answer:

1. 

2. Field lines of the electrostatic field have the following properties:

• Never intersect each other.
• Electrostatic field lines never form closed loops.

Question 3. A system has two charges q_A = 2.5 x 10^-7C and q_B = – 2.5 x 10^-7C located at points A : (0, 0, -15 cm) and B : (0, 0, +15 cm), respectively. What is the total charge and electric dipole moment of the system?

Answer:

Total charge = 2.5 x 10^-7– 2.5 x 10^-7 = 0

Electric dipole moment is \(\overrightarrow{\mathrm{p}}=\mathrm{q}(2 \overrightarrow{\mathrm{a}})\)

= 2.5×10^-7x (0.15+0.15) C-m

= 7.5×10⁸ C-m

The direction of the dipole moment is along the Z-axis.

Question 4. Find the expression for torque experienced by an electric dipole in a uniform electric field.

Answer:

Torque on an electric dipole in a uniform electric field

We consider a dipole with charges +q and -q which are at a distance d away from each other. Let it be placed in a uniform electric field of strength E such that the axis of the dipole forms an angle 0 with the electric field.

The force on the charges is

F_+q = +qE → towards the direction of the electric field

F_-q = -qE → opposite to the direction of the electric field

Since the magnitudes of forces are equal and they are separated by a distance d,

The torque on the dipole is given by :

Torque (τ) = Force x perpendicular distance between both forces

τ = F.d sin 0

or τ = qEdsinO

So τ = pE .sin 0 (p = qd)

or \(\vec{\tau}=\vec{p} \times \vec{E}\) [in vector form]

Question 5.

1. Define electric flux Write its SI Unit.
2. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?

Answer:

1. Electric flux is proportional to the number of electric field lines passing through a given area.

It is numerically equal to the dot product of the electric field and area vector.

Φ = A (Ecosθ)

⇒ \(\phi={\vec{E}} \cdot {\vec{A}}\)

The electric flux through an area is the dot product of the magnitude of \(\vec{E}\) and \(\vec{A}\).

The S.I. unit of ‘electric flux’ is N-m²C^-1 or V-m.

2. Wc knows that flux through the Gaussian surface is given by \(\phi=q / \varepsilon_0\)

As flux is independent of radius, it is not affected by changing the radius.

Question 6.

1. A uniformly charged large plane sheet has charge density \(\sigma=\left(\frac{1}{18 \pi}\right) \times 10^{-15} \mathrm{C} / \mathrm{m}^2\). Find the electric field at point A which is 50 cm from the sheet. Consider a straight line with three points P. Q and R, placed 50 cm from the charged sheet on the right side as shown in the figure. At which of these points, does the magnitude of the electric field due to the sheet remain the same as that at point A and why?

2. Two small identical conducting spheres carrying charge 10 μC and- 20μC when separated by a distance r, experience a force F each. If they are brought in contact and then separated to a distance of \(\frac{r}{2}\), what is the new force between them in terms of F?

Answer:

1. \(E=\frac{\sigma}{2 \varepsilon_0}=\frac{1}{18 \pi} \times \frac{10^{-15}}{2 \times 1} \times 4 \pi \mathrm{k}\)  (\(k=\frac{1}{4 \pi \varepsilon_0}\))

= \(\frac{1}{9} \times 10^{-15} \times 9 \times 10^9\)

= 10^-6 V/m

Point →Q, Because at 50 cm, the charge sheet acts as a finite sheet, and thus the magnitude
remains the same towards the middle region of the planar sheet.

2.

According to Coulomb’s Law

⇒ \(\mathrm{F}=\frac{\mathrm{k}(10 \mu \mathrm{C})(20 \mu \mathrm{C})}{\mathrm{r}^2}\)

⇒ \(\mathrm{F}=\frac{2 \times 9 \times 10^9 \times 100 \times 10^{-12}}{\mathrm{r}^2}\)

⇒ \(F=\frac{1.8}{r^2}\) →(1)

After contact

⇒ \(F^{\prime}=\frac{k(-.5 \mu C)(-.5 \mu C)}{\left(\begin{array}{l}
r{\prime} \\
2\end{array}\right)^2}\)

⇒ \(F^{\prime}=\frac{4 \times 9 \times 10^{11} \times 2.5 \times 10^{-12}}{r^2}\)

∴ \(F^{\prime}=\frac{0.9}{r^2}\) →(2)

So, \(F^{\prime}=\frac{F}{2}\)

Question 7. A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity \(\vec{E}\) at a point on the axis of the ring. Hence shows that for points at a large distance from the ring, it behaves like a point charge.

Answer:

Suppose that the ring is placed with its plane perpendicular to the x-axis. as shown in the above diagram. We consider a small element dl of the ring. So the charge dq on the element dl is

⇒ \(\mathrm{dq}=\frac{\mathrm{q}}{2 \pi \mathrm{a}} \mathrm{d} l\) [∵ \(\lambda=\frac{\mathrm{q}}{2 \pi \mathrm{a}}\)] = charge per unit length

∴ The magnitude of the field de has two components:

1. The axial component ⇒ dEcosθ
2. The perpendicular component ⇒ dEsinθ

Since the perpendicular components of any two diametrically opposite elements are equal and opposite, they all cancel out in pairs. Only the axial components will add up to produce the resultant field at point P, which is given by,

⇒ \(\mathrm{E}=\int_0^{2 \pi \mathrm{a}} \mathrm{d} \mathrm{E} \cos \theta\)

⇒ \(\mathrm{E}=\int_0^{2 \pi \mathrm{a}} \frac{\mathrm{kq}}{2 \pi \mathrm{a}} \frac{\mathrm{d}
l}{\mathrm{r}^2}\left(\frac{\mathrm{x}}{\mathrm{r}}\right)=\frac{\mathrm{kqx}}{2 \pi \mathrm{ar}^3} \int_0^{2 \pi \mathrm{a}} \mathrm{d} l\) [∵ \(\cos \theta=\frac{x}{r}\)]

∴ \(\mathrm{E}=\frac{\mathrm{kqx}}{2 \pi \mathrm{ar}^3}[l]_0^{2 \pi \mathrm{a}}=\frac{\mathrm{kqx}}{2 \pi \mathrm{a}} \frac{1}{\left(\mathrm{x}^2+\mathrm{a}^2\right)^{3 / 2}}(2 \pi \mathrm{a}) \quad\left[\mathrm{as}^2=\mathrm{x}^2+\mathrm{a}^2\right]\)

or \(\mathrm{E}=\frac{\mathrm{kqx}}{\left(\mathrm{x}^2+\mathrm{a}^2\right)^{3 / 2}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{qx}}{\left(\mathrm{x}^2+\mathrm{a}^2\right)^{3 / 2}}\)

Special case: For points at large distance from the ring x >> a

∴ \(E=\frac{k q}{x^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q}{x^2}\)

Tins are the same as the field due to a point charge, indicating that for far-off axial points, the charged ring behaves as a point charge.

Question 8. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 x 10^-22 C/m². What is electric field intensity:

1. In the outer region of the first plate.
2. In the outer region of the second plate, and
3. Between the plates?

Answer:

Where. E_P = Electric field due to Plate P

E_Q = Electric field due to Plate Q

1. In the outer region of the first plate

⇒ \(E_1=E_P-E_Q=\frac{\sigma_P-\sigma_Q}{2 \varepsilon_0}\)

= \(\frac{17 \times 10^{-22}-17 \times 10^{-22}}{2 \varepsilon_0}\) = E₁ = 0 (i.e Electric field is zero)

2. Similarly, the electric field is zero in this case also E₁₁ = E_P– E_Q = 0

3. Between the plates

⇒ \(\mathrm{E}_{\mathrm{3}}=\mathrm{E}_{\mathrm{P}}+\mathrm{E}_{\mathrm{Q}}=\frac{\sigma_{\mathrm{P}}+\sigma_{\mathrm{Q}}}{2 \varepsilon_0}\)

= \(\frac{17 \times 10^{-22}+17 \times 10^{-22}}{2 \varepsilon_0}\)

∴ \(\mathrm{E}_{3}=\frac{34 \times 10^{-22}}{2 \times 8.854 \times 10^{-12}}=1.92 \times 10^{-10} \mathrm{NC}^{-1}\)

Question 9. Derive an expression for the electric field intensity at a point on the equatorial line of an electric dipole of dipole moment \(\vec{P}\) and length 2a. What is the direction of this field?

Answer:

We consider a dipole consisting of -q and +q separated by a distance 2a. Let P be a point on the equatorial line.

⇒ \({\vec{E}}_{\mathrm{A}}=\frac{\mathrm{l}}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{(\mathrm{AP})^2} \text { along } \vec{PA}\)

⇒ \(E_A=\frac{1}{4 \pi \varepsilon_0} \frac{q}{\left(r^2+a^2\right)}\)

⇒ \({\vec{E}}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{(\mathrm{BP})^2} \text { along } \vec{BP}\)

∴ \(E_B=\frac{1}{4 \pi \varepsilon_0} \frac{q}{\left(r^2+a^2\right)}\)

The resultant intensity is the vector sum of the intensities along PA and BP. E_A and E_B can be resolved into vertical and horizontal components. The vertical components of E_A and E_B cancel each other as they are equal and oppositely directed. So the horizontal components add up to the resultant field.

E = E_A cos θ + E_B cos θ

E = 2E_Acosθ , as E_A = E_B

Substituting, \(\cos \theta=\frac{a}{\left(r^2+a^2\right)^{\frac{1}{2}}}\) in the above equation

⇒ \(E=2 E_A \cos \theta=\frac{2}{4 \pi \varepsilon_0} \frac{q}{\left(r^2+a^2\right)} \frac{a}{\left(r^2+a^2\right)^{\frac{1}{2}}}\)

∴ \(\mathrm{E}=\frac{\mathrm{kp}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}}\)along \(\overrightarrow{\mathrm{BA}}\) (As p+qx2a)

As a special case,

If a²<<r² then, \(E=\frac{k p}{r^3}\) along \(\overrightarrow{\mathrm{BA}}\)

Electric field intensity at an axial point is twice the electric field intensity on the equatorial line.

The direction of the field will be against the direction of the dipole moment.

Question 10. Four point charges Q, q, Q, and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the resultant electric force on a charge Q

Answer:

Let us find the force on the charge Q at the point C. Force due to the other charge Q

∴ \(\mathrm{F}_1=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}^2}{(\mathrm{a} \sqrt{2})^2}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{\mathrm{Q}^2}{2 \mathrm{a}^2}\right)\) (along AC)

Force due to the charge q placed at B

⇒ \(\mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{qQ}}{\mathrm{a}^2} \text { along } \mathrm{BC}\)

Force due to the charge q placed at D

⇒ \(F_3=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{qQ}}{\mathrm{a}^2} \text { along } \mathrm{DC}\) along DC

Resultant of these two equal forces F₂ and F₃

⇒ \(\mathrm{F}_{23}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{qQ}(\sqrt{2})}{\mathrm{a}^2} \text { (along } \mathrm{AC} \text { ) }\)

∴ The net force on charge Q (at point C)

⇒ \(F=F_1+F_{23}=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{\mathrm{a}^2}\left[\frac{Q}{2}+\sqrt{2} \mathrm{q}\right]\)

This force is directed along the AC

(For the charge Q, at the point A, the force will have the same magnitude but will be directed along CA)

Long Questions And Answers

Question 1. When an electric dipole is placed in a uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on the electric dipole in a uniform electric field is zero. However, these forces are not collinear, so they give rise to some torque on the dipole. Since the net force on an electric dipole in a uniform electric field is zero. So no work is done in moving the electric dipole in a uniform electric field. However, some work is done in rotating the dipole against the torque acting on it.

1. The dipole moment of a dipole in a uniform external field \(\vec{E}\) is \(\vec{p}\). Write the expression of torque acting on the dipole.
2. An electric dipole consists of two opposite charges, each of magnitude 1.0 μC separated by a distance of 2.0 cm. The dipole is placed in an external field of 10⁵ NC^-1. Find the value of max Torque.
3. Write the value of angle θ, when τ is minimum.
4. When an electric dipole is held at an angle θ (θ ≠ 0° or 180°) in a uniform electric field Write the value of net force \(\vec{F}\) and torque \(\vec{\tau}\)?

Answer:

1. \(\vec{\tau}=\vec{p} \times \vec{E}\)

2. \(\tau=P E \sin 90^{\circ}=p E=1 \times 10^{-6} \times 2 \times 10^{-2} \times 10^5=10^{-3} \mathrm{~N} . \mathrm{m}\)

3. 0° or 180° or nπ

4. F = 0, τ ≠ 0

Question 2. Concept of Electric field

An electric field is an elegant way of characterizing the electrical environment of a system of charges. An electric field at a point in the space around a system of charges tells you the force a unit-positive test charge would experience if placed at that point (without disturbing the system). The electric field is a characteristic of the system of charges and is independent of the last charge that you place at a point to determine the field

1. Write one property of electric field lines
2. Define electric field intensity.
3. The SI unit of the electric field is_______
4. A proton of mass ‘m’ placed in the electric field region remains stationary in the air. What is the magnitude of the electric field?

Answer:

1. The electric field line starts from +ve charge and ends at -ve charge.
2. It is defined as the electric force experienced per unit positive test charge is known as electric field intensity.
3. N/C and V/m
4. mg = eE
   E = mg/c

Question 3.

1. Use Gauss’ law to derive the expression for the electric field \((\vec{E})\) due to a straight uniformly charged infinite line of charge density λ C/m.
2. Draw a graph to show the variation of E with perpendicular distance r from the line of
   charge.
3. Find the work done in bringing a charge q at a perpendicular distance from co-long charged
   wire r₁, to r₂, (r₂ > r₁).

Answer:

1. To calculate the electric field, imagine a cylindrical Gaussian surface, since the field is everywhere radial, flux through two ends of the cylindrical Gaussian surface is zero. At the cylindrical part of the surface electric field E is normal to the surface at every point and its magnitude is constant. Therefore flux through the Gaussian surface.

= Flux through the curved cylindrical part of the surface.

= E x 2πl → (1)

Applying Gauss’s Law

⇒ \(\text { Flux } \phi=\frac{q_{\text {enclosed }}}{\varepsilon_0}\)

Total charge enclosed

= Linear charge density x l = xl

∴ \(\phi=\frac{\lambda l}{\varepsilon_0}\) → (2)

Using Equations (1) and (2)

∴ \(\mathrm{E} \times 2 \pi \mathrm{r} l=\frac{\lambda l}{\varepsilon_0} \Rightarrow \overrightarrow{\mathrm{E}}=\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}} \hat{\mathrm{n}}\)

(where \(\hat{n}\) is a unit vector normal to the line charge)

2. The required graph is as shown :

3. Work done in moving a charge q with displacement ‘dr’

⇒ \(\mathrm{dW}=\overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}\)

⇒ \(\mathrm{dW}=\mathrm{q} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}=\mathrm{q} \mathrm{Edr} \cos θ\)

⇒ \(\mathrm{dW}=\mathrm{q} \times \frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}} \mathrm{dr} \)

Work done in moving the given charge from r₁ to r₂ (r₂ > r₁)

⇒ \(\int_{r_1}^{r_2} d W=\int_{r_1}^{r_2} \frac{\lambda q d r}{2 \pi \varepsilon_0 r}\)

⇒ \(\mathrm{W}=\frac{\lambda \mathrm{q}}{2 \pi \varepsilon_0}\left[\log _{\mathrm{e}} \mathrm{r}_2-\log _{\mathrm{e}} \mathrm{r}_1\right]\)

∴ \(\mathrm{W}=\frac{\lambda \mathrm{q}}{2 \pi \varepsilon_0}\left[\log _{\mathrm{e}} \frac{\mathrm{r}_2}{\mathrm{r}_1}\right]\)

Question 4. Define electric flux. Is it a scalar or a vector quantity?

1. A point charge q is at a distance of d/2 directly above the center of a square of side d, as shown in the figure. Use Gauss’ law to obtain the expression for the electric flux through the square.

2. If the point charge is now moved to a distance ‘d’ from the center of the square and the side of the square is doubled, explain how the electric flux will be affected.

Answer:

1. Electric flux through a given surface is defined as the dot product of the electric field and area vector over that surface.

Alternatively \(\phi=\int \vec{E} \cdot d \vec{S}\)

Also, accept

Electric flux, through a surface, equals the surface integral of the electric field over a closed surface. It is a scalar quantity.

Constructing a cube of side ‘d’ so that charge ‘q’ is placed within this cube (Gaussian surface)

According to Gauss’ law the Electric flux \(\phi=\frac{\text { charge enclosed }}{\varepsilon_0}=\frac{\mathrm{q}}{\varepsilon_0}\)

This is the total flux. through all the six faces of the cube.

Hence electric flux through the square \(\frac{1}{6} \times \frac{q}{\varepsilon_0}=\frac{q}{6 \varepsilon_0}\)

2. If the charge is moved to distance d and the side of the square is doubled even then the total charge enclosed in it will remain the same. Hence the total flux will remain the same as before.

Question 5.

1. Derive an expression for the electric field E due to a dipole of length ‘2a’ at a point distant r from the center of the dipole on the axial line.
2. Draw a graph of E versus r for r >> a.
3. If this dipole were kept in a uniform external electric field E₀, with the help of a diagram represent the position of the dipole in stable and unstable equilibrium and write the
   expressions for the torque acting on the dipole in both cases.

Answer:

1. Let’s consider a dipole system,

Here, AO = OB = a

OP = r

BP = r-a

AP = r + a

Elec, field (\(\overrightarrow{\mathrm{E}}_{\mathrm{B}}\)), due to the charge at point ‘B’ being towards ‘P’

Elec, field (\(\overrightarrow{\mathrm{E}}_{\mathrm{A}}\)), due to the charge at point ‘A’ being opposite to ‘P’

Now, according to the superposition principle,

⇒ \(\overrightarrow{\mathrm{E}}_{\mathrm{p}}=\overrightarrow{\mathrm{E}}_{\mathrm{ix} \mathrm{axial}}=\overrightarrow{\mathrm{E}}_{\Lambda}+\overrightarrow{\mathrm{E}}_{\mathrm{B}}\)

E_P = E_B-E_A

⇒ \(E_p=\frac{-k q}{(r+a)^2}+\frac{k q}{(r-a)^2}=k q\left[\frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}\right]=k q\left[\frac{(r+a+r-a)(r+a-r+a)}{(r-a)^2(r+a)^2}\right]\)

=\(\frac{k q(2 r)(2 a)}{\left(r^2-a^2\right)^2}\)

⇒ \(E_{a x i a l}=\frac{2 r k(2 a)(q)}{\left(r^2-a^2\right)^2}\) [∵ \(|\overrightarrow{\mathrm{p}}|=2 \mathrm{a} \times \mathrm{q}\)] = \(\frac{2 \mathrm{kpr}}{r^{4}}\)

if a₂ << r₂ then \(\dot{\vec{E}}_{\mathrm{axial}}=\frac{2 \mathrm{k} {\vec{p}}}{\mathrm{r}^3}\)

It will be directed in the direction of electric dipole moment, \(\vec{P}\).

2.

3. (1). Stable equilibrium

Torque (τ) = pEsinθ

= pE x sinθ° = 0 (vsinθ° = 0)

(2). Unstable equilibrium

Torque (τ) = pE sinθ

= pE sin 180°

= pE x 0 = 0 (∵ sin 180° = 0)

Question 6.

1. Use Gauss’ theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density σ.
2. An infinitely large thin plane sheet has a uniform surface charge density +σ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point distant r, in front of the charged plane sheet.

Answer:

1.

As shown in the figure, considering a cylindrical Gaussian surface of cross-section A

Flux through the curved surface :

∴ \(\phi=\int \overrightarrow{\vec{E}} \cdot \mathrm{d} {\vec{S}}=\int \mathrm{Eds} \cos 90^{\circ}=0\)

At the points on the curved surface, the field vector E and area vector dS make an angle of 90° with each other. Therefore, curved surfaces do not contribute to the flux.

Flux through end caps :

∴ \(\phi=\oint \vec{E} \cdot \mathrm{dS}=\oint \mathrm{EdS} \cos 0^{\circ}=\mathrm{EA}\)

Hence, the total flux through the closed surface is :

Φ = Flux through both end caps + flux through curved surface

or Φ = EA + EA + 0 = 2EA → (1)

Now according to Gauss’ law for electrostatics

⇒ \(\phi=q / \varepsilon_0\) → (2)

Comparing equations (1) and (2), we get

2EA = q/ε₀

E = q/2ε₀A → (3)

The area of the sheet enclosed in the Gaussian cylinder is also A. Therefore, the charge contained in the cylinder, q = σA as a (surface charge density) = q/A.

Substituting this value of q in equation (3), we get

E = σA/2ε₀A

or E = σ/2ε₀

This is the relation for the electric field due to an infinite plane sheet of charge. The field is uniform and does not depend on the distance from the plane sheet of charge.

2. \(V=\frac{W}{c}=\int_{\infty}^r \vec{E} \cdot d \vec{r}\)

⇒ \(\mathrm{W}=\mathrm{c} \int_{\infty}^{\mathrm{r}}(-\mathrm{Edr})\)

⇒ \(W=-q \int_{\infty}^r \frac{\sigma}{2 c_0} d r\)

∴ \(W=\frac{q \sigma}{2 \varepsilon_0}(\infty-r)\) W = ∞

Electrostatic Potential And Capacitance Multiple Choice Questions

Question 1. The electric potential a; a point on the axis of a short electric dipole, at a distance x from the midpoint of the dipole is proportional to

1. \(\frac{1}{x}\)
2. \(\frac{1}{x^3/2}\)
3. \(\frac{1}{y^3}\)
4. \(\frac{1}{x^2}\)

Answer: 4. \(\frac{1}{x^2}\)

V = \(\frac{k p \cos θ}{r^2}\)

Question 2. In the given network all capacitors used are identical and each one is of capacitance C. Which of the following is the equivalent capacitance between the points A and B?

1. 6 C
2. 5/2 C
3. 3/2 C
4. 5/6 C

Answer: 3. 3/2 C

Question 3. A particle having a mass 1 g and an electric charge 10^-8 C travels from point A having an electric potential of 600 V to point B having zero potential. What would be the change in its kinetic energy?

1. \(-6 \times 10^{-6} \mathrm{erg}\)
2. \(-6 \times 10^{-6} \mathrm{~J}\)
3. \(6 \times 10^{-6} \mathrm{~J}\)
4. \(6 \times 10^{-6} \mathrm{crg}\)

Answer: 3. \(6 \times 10^{-6} \mathrm{~J}\)

ΔK = \(\mathrm{W}=\mathrm{q}\left(\mathrm{V}_1-\mathrm{V}_2\right)\)

= \(10^{-8}(600-0)\)

= \(6 \times 10^{-6} \mathrm{~J}\)

Question 4. The energy of a charged capacitor is U, Now it is removed from a battery and then is connected to another identical uncharged capacitor in parallel. What will be the energy of each capacitor now,

1. 3U/2
2. U
3. U/4
4. U/2

Answer: 3. U/4

U, Q, C, V

U = \(\frac{Q^2}{2C}\)….(1)

Now,

In parallel, the charge will same equal to \(\frac{Q}{2}\) on each capacitor.

So, \(U^{\prime}=\frac{Q^2}{2 C}=\frac{\left(\begin{array}{c}\mathrm{Q} \\ 2\end{array}\right)^2}{2 \mathrm{C}}=\frac{\mathrm{Q}^2}{8 \mathrm{C}}\)…(2)

So, \(\frac{U^{\prime}}{U}=\frac{1}{8} \times 2=\frac{1}{4}\)

⇒ \(U^{\prime}=\frac{U}{4}\)

Question 5. The capacitance of a variable capacitor joined with the batten of 100 V is changed from 2μF to 10μF. What is the change in the energy stored in it?

1. \(2 \times 10^{-2} \mathrm{~J}\)
2. \(2.5 \times 10^{-2} \mathrm{~J}\)
3. \(6.5 \times 10^{-2} \mathrm{~J}\)
4. \(4 \times 10^{-2} \mathrm{~J}\)

Answer: 4. \(4 \times 10^{-2} \mathrm{~J}\)

U = \(\frac{1}{2} \mathrm{CV}^2\)

Δ U =\(\frac{1}{2} C_2 V^2-\frac{1}{2} C_1 V^2\)

= \(\frac{1}{2} \times 100 \times 100(10-2) \times 10^{-6}=4 \times 10^{-2} J\)

Question 6. The distance between electric charges Q C and 9 Q C is 4 m. What is the electric potential at a point of the line joining them where the electric field is zero?

1. 4 kQ V
2. 10 kQ V
3. 2 kQ V
4. 2.5 kQ V

Answer: 1. 4 kQ V

The electric field at P is zero \(\frac{k Q}{x^2}=\frac{k(9 Q)}{(4-x)^2}\)

⇒ \(9 x^2=(4-x)^2 \Rightarrow 3 x=4-x\)

x=1 m

So \(V=\frac{k Q}{(1)}+\frac{k(9 Q)}{3}=k Q+3 k Q=4 k Q\) volt

Question 7. For a capacitor, the distance between two plates is 5v and the electric field between them is E₀. Now a dielectric slab having dielectric constant 3 and thickness v is placed between them in contact with one plate. In this condition, the potential difference between the two plates is

1. \(15 \mathrm{E}_0 \mathrm{X}\)
2. \(7 E_0 x\)
3. \(\frac{13 E_0 x}{3}\)
4. \(\frac{9 E_0 \mathrm{x}}{2}\)

Answer: 3. \(\frac{13 E_0 x}{3}\)

V = \(E_0(5 x-x)+E_m \times x\)

= \(E_0(4 x)+\frac{E_{01}}{3} x\left[E_m=\frac{E_0}{c_t}\right]\)

⇒ \(\frac{13 E_0 x}{3}\)

Question 8. A Point P is 40 m away from the 20 μC point charge and 20 m from the 4 μC point charge. The electric potential al P is V. (k = 9 x 10⁹ Nm² C^-2]

1. 1300
2. 6300
3. 2700
4. 4500

Answer: 2. 6300

V = \(\frac{k \times 20 \mu \mathrm{C}}{40}+\frac{k \times 4 \mu \mathrm{C}}{20} \Rightarrow \frac{28 \mathrm{k}}{40} \mu \mathrm{C} \)

= \(\frac{7}{10} \times 9 \times 10^9 \times 10^{-6}\)

V = \(6300 \text { volt }\)

Question 9. If a capacitor having a capacitance of 1200 μF is charged at a uniform rate of 100 μC/s, what is the time required to increase its potential by 20 volts?

1. 500 s
2. 6000 s
3. 240 s
4. 120 s

Answer: 3. 240 s

q = C V

V = \(\frac{q}{C}\)

⇒ \(\frac{\Delta V}{\Delta t}=\frac{1}{C} \frac{\Delta q}{\Delta t}\)

⇒ \(\frac{20}{t}=\frac{1}{1200 \mu F} \times 100 \frac{\mu C}{s}\)

t = \(\frac{1200 \times 20}{100}=240 \mathrm{sec}\)

Question 10. For uniform electric field \(\vec{E}=E_0(\hat{j})\), if the electric potential at y = 0 is zero, then the value of electric potential at y = +y will the

1. -yE₀
2. yE₀
3. y²E₀
4. -y²E₀

Answer: 1. -yE₀

⇒ \(\int \mathrm{dV}=-\int \mathrm{Edr}\)

⇒ \(\int_0^1 \mathrm{dV}=-\mathrm{E}_0 \int_0^\gamma \mathrm{dy}\)

⇒ \(\mathrm{V}=-\mathrm{E}_0 \mathrm{y}\)

Question 11. Equipolcnlial Surface through a point is _____ to the electric field at that point.

1. Normal
2. Parallel
3. At an angle of 45°
4. At an angle of 30°

Answer: 1. Normal

Question 12. A particle having charge ‘q’ is accelerated with a potential difference ΔV it would gain energy of______

1. q ΔV²
2. q²ΔV
3. qΔV
4. q²ΔV²

Answer: 3. qΔV

Question 13. In a parallel plate capacitor, the area of each plate A 1 m² and the distance between two plates d = 1 mm. Then the capacitance of a capacitor C = ____ F.

1. 8.85 x 10^-12
2. 8.85 x 10^-9
3. 8.85 x 10^-6
4. 8.85 x 10^-15

Answer: 2. 8.85 x 10^-9

C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)

= \(\frac{8.85 \times 10^{-12} \times 1}{1 \times 10^{-3}}=8.85 \times 10^{-9} \mathrm{~F}\)

Electrostatic Potential And Capacitance Assertion And Reason Type Questions And Answers

For question numbers 14 to 18 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1), (2), (3), and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: Electrons move away from a low potential to a high potential region.

Reason: Because electrons have a negative charge.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 2. Assertion: The capacity of a given conductor remains the same even if the charge is varied on it.

Reason: Capacitance depends upon nearly medium as well as the size and shape of the conductor.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 3. Assertion: A conductor having equal positive charge and volume, must also have the same potential.

Reason: Potential depends only on the charge and volume of the conductor.

Answer: 4. A is false and R is also false

Question 4. Assertion: A metallic shield in the form of a hollow shell may be built to block an electric field.

Reason: In a hollow spherical shell, the electric field inside it is zero at every point.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 5. Assertion: When charges are shared between any two bodies no charge is really lost, some loss of energy does occur.

Reason: Some energy disappears in the form of heat.

Answer: 1. Both A and R are true and R is the correct explanation of A

Electrostatic Potential And Capacitance Short Answer Type Questions

Question 1. A 600 pF capacitor is charged by a 200V supply. It is then disconnected from the supply and connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Answer:

Energy loss

ΔU = \(\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2\)

= \(\frac{1}{2} \frac{600 \times 600 \times 10^{-24}}{(600+600) \times 10^{12}}(200-0)^2\)

= \(\frac{1}{2} \times \frac{60(0 \times 600}{1200} \times 10^{-12} \times 4 \times 10^4\)

= \(6 \times 10^{-6} \mathrm{~J}\)

Question 2. A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?
Answer:

C = \(\frac{\varepsilon_0 A}{(d-1)+\frac{1}{K}}=\frac{\varepsilon_0 A}{\left(d-\frac{3}{4} d\right)+\frac{3 d}{4 K}}=\frac{\varepsilon_0 A}{d\left(1+\frac{3}{K}\right)}\)

Here \(\left(C_0=\frac{\varepsilon_0 A}{d}\right)\)

C = \(\frac{4 K C_0}{K+3}\)

Question 3. Four point charges Q, q. Q and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the potential energy of this system. 

Answer:

The potential energy of the system

U = \(\frac{1}{4 \pi \varepsilon_0}\left[4 \frac{q Q}{a}+\frac{q^2}{a \sqrt{2}}+\frac{Q^2}{a \sqrt{2}}\right]\)

= \(\frac{1}{4 \pi \varepsilon_0 a}\left[4 q Q+\frac{q^2}{\sqrt{2}}+\frac{Q^2}{\sqrt{2}}\right]\)

Question 4. Two charges 3 x 10^-8 C and -2 x 10^-8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to he zero.
Answer:

Let at point P potential is zero

V_p = V₁ + V₂= 0

⇒ \(V_F=\frac{k q_1}{x}+\frac{k q_2}{(15-x)}=0\)

⇒ \(V_f=k\left[\frac{3 \times 10^{-k}}{x}-\frac{2 \times 10^{-x}}{(15-x)}\right]=0\)

⇒ \(\frac{3 \times 10^{-x}}{x}=\frac{2 \times 10^{-x}}{(15-x)}\)

⇒ 45 – 3x = 2x

⇒ x = 9cm

Question 4. Three capacitors of capacitance C₁, C₂, and C₃ are connected in series to a source of V volt. Show that the total energy stored in the combination of capacitors is equal to the sum of the energy stored in individual capacitors.
Answer:

So. w.k.t. electric energy stored in the capacitor

U = \(\frac{\mathrm{Q}^2}{2 \mathrm{C}}\) (Series combination Q → constant)

⇒ \(\mathrm{U}_1=\frac{\mathrm{Q}^2}{2 \mathrm{C}_1} \cdot \mathrm{U}_2=\frac{\mathrm{Q}^2}{2 \mathrm{C}_2} \cdot \mathrm{U}_3=\frac{\mathrm{Q}^2}{2 \mathrm{C}_3}\)

So, Total energy \(\mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_3=\frac{\mathrm{Q}^2}{2}\left(\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_2}\right)\)

In Series Combination w.k.t. \(\frac{1}{C_{\text {eq}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)

So, \(U_1+U_2+U_3=\frac{Q^2}{2 C_{\text {eq}}}\)

⇒ \(\mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_1=\mathrm{U}_{1 \text { Total }}\)

Question 5. Two identical parallel plate capacitors A and B are connected to a batter of V volts with switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Land the ratio of the total electrostatic energy stored in both capacitors before and alter the introduction of the dielectric.

Answer:

When switch ‘S’ is closed, the initial energy (E_i) of the system, \(\mathrm{E}_{\mathrm{i}}=\frac{1}{2} \mathrm{CV}^2+\frac{1}{2} \mathrm{CV}^2=\mathrm{CV}^2\)

If \(\frac{1}{2} \mathrm{CV}^2=\mathrm{U} \text {. then } \mathrm{E}_{\mathrm{i}}=\mathrm{CV}^2=2 \mathrm{U}\)….(1)

When switch ‘S’ is opened then capacitor ‘A’ gets a constant supply of voltage (V) and on the other hand, charge ‘Q’ becomes constant in capacitor ‘B’.

∴ In the above situation, if the dielectric of strength ‘K’ is filled, we may write.

The final energy of the system as \(\mathrm{E}_{\mathrm{f}}=\frac{1}{2}(\mathrm{KC}) \mathrm{V}^2+\frac{1}{2}\left(\frac{\mathrm{Q}^2}{\mathrm{KC}}\right)\)

or, \(\mathrm{E}_{\mathrm{f}}=\mathrm{KU}+\frac{\mathrm{U}}{\mathrm{K}}=\mathrm{U}\left(\frac{\mathrm{K}^2+1}{\mathrm{~K}}\right)\)

because \(\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}=\mathrm{U}\)

Now, \(\frac{\mathrm{E}_{\mathrm{i}}}{\mathrm{E}_{\mathrm{f}}}=\frac{2 \mathrm{U}(\mathrm{K})}{\mathrm{U}\left(\mathrm{K}^2+1\right)}=\frac{2 \mathrm{~K}}{\mathrm{~K}^2+1}\)

Question 6. Define an equipotential surface. Draw equipotential surfaces

1. In the case of a single-point charge.
2. In a constant electric field in Z-direction.
3. Can an electric field exist tangential to an equipotential surface? Give reason.

Answer:

An equipotential surface is a surface that has equal potential at every point it. The net work done in moving a charge from one point to another on this surface is zero.

1. A point charge

2.

3. A number of electric fields cannot exist tangential to an equipotential surface. If it happens then a charged particle will experience a force along the tangential line and can move along it. Since a charged particle can move only due to the potential difference, this contradicts the concept of an equipotential surface.

Question 7. Obtain an expression for the work done to dissociate the system of three charges placed at the vertices of an equilateral triangle of side ‘a’ as shown below.

Answer:

The total electrostatic potential energy of the system

U = \(U_{A B}+U_{B C^C}+U_{C A}=k\left[\frac{q(-4 q)}{a}+\frac{(-4 q)(2 q)}{a}+\frac{q(2 q)}{a}\right]=-k \times \frac{10 q^2}{a}\)

∴ Work done to dissociate the system W = -U

W = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{10 q^2}{a}\)

Question 8. A 200 μF parallel plate capacitor having a plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Find

1. Capacitance,
2. The electric field between the plates,
3. The energy density of the capacitor will change?

Answer:

As we know for a capacitor C = \(\frac{\varepsilon_0 \mathrm{~A}}{d}\)

and for a capacitor partially filled with dielectric, capacity is C = \(\frac{\varepsilon_0 A}{(d-1)+\frac{1}{K}}\)

here, we may consider the term [(d – t) + \(\frac{1}{K}\)] as effective distance, which can be calculated for the present situation as,

⇒ \(\mathrm{d}_{\mathrm{eff}}=(2 \mathrm{~d}-\mathrm{t})+\frac{\mathrm{t}}{\mathrm{K}}=2 \times 5-5+\frac{5}{10}=5.5 \mathrm{~mm}\)

Thus,

1. Effective new capacitance = \(200 \mu \mathrm{F} \times \frac{5 \mathrm{~mm}}{5.5 \mathrm{~mm}}=\frac{2000}{11} \mu \mathrm{F}\)

= \(182 \mu \mathrm{F} \text { as } \rightarrow \frac{\mathrm{C}_{\text {eff }}}{\mathrm{C}}=\frac{\mathrm{d}}{\mathrm{d}_{\text {eff }}}=\frac{\mathrm{E}_{\text {eff }}}{\mathrm{E}}\)

2. Effective new electric field = \(\frac{100}{5.5 \times 10^{-3} \mathrm{~m}}=\frac{200000}{11} \simeq 18182 \mathrm{~V} / \mathrm{m}\)

3. \(\frac{\text { New energy Stored }}{\text { Original energy Stored }}=\frac{{ }_2^1 \mathrm{C}_{\mathrm{eff}} \mathrm{V}^2}{{ }_2^1 \mathrm{CV}^2}=\frac{\mathrm{C}_{\mathrm{eff}}}{\mathrm{C}}=\frac{10}{1 \mathrm{I}}\)

Thus, the new energy density will be (10/11)² of the original energy density = 100/121 times

Question 9. In a network, four capacitors C₁, C₂, C₃, and C₄ are connected as shown

1. Calculate the net capacitance in the circuit.
2. If the charge on the capacitor C₁ is 6 μC,
   1. Calculate the charge on the capacitors C₃ and C₄, and
   2. Net energy is stored in the capacitors C₃ and C₄ connected in series.

Answer:

1. Equivalent capacitance of C₁ and C₂

C” = C₁ + C₂

C’ = 9μF

Equivalent capacitance of the circuit \(\frac{q_3^2}{2 C_3}+\frac{q_4^2}{2 C_4} \quad\left\{q_4=q_3\right\}\)

U = \(\frac{q_3^2}{2}\left[\frac{1}{C_3}+\frac{1}{C_4}\right]=\frac{18 \times 18}{2}\left[\frac{C_4+C_3}{C_3 \times C_4}\right]\)

U = \(\frac{18 \times 18}{2}\left[\frac{16}{12 \times 4}\right] \Rightarrow U=54 \mu \mathrm{J}\)

Question 10. Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of 10 V is connected across A and B. calculate the charge drawn from the battery.

Answer:

The above circuit is a balanced Wheatstone bridge. So there is no flow of current in 50 uμF capacitor (between P and R). so this capacitor will not play any role in the circuit.

∴ modified circuit is ⇒

C₁ ,C₂ are in scries \(\frac{1}{C^{\prime}}=\frac{1}{10 \mu \mathrm{F}}+\frac{1}{20 \mu \mathrm{F}} \Rightarrow \mathrm{C}^{\prime}=\frac{20}{3} \mu \mathrm{F}\)

C₃, C₄ are in series

Now, \(C^{\prime \prime}=\frac{1}{5 \mu \mathrm{F}}+\frac{1}{10 \mu \mathrm{F}} \Rightarrow \mathrm{C}^{\prime \prime}=\frac{10}{3} \mu \mathrm{F}\)

Equivalent capacitance of the circuits C” + C”

⇒ \(C_{\mathrm{eq}}=\frac{20}{3}+\frac{10}{3}=10 \mu \mathrm{F}\)

Now, charge drawn from the battery

V = 10 volt (given)

Q = C_eq V ⇒ Q = 10 x 10 x 10^-6= 10^-4 C

Question 11. In the following arrangement of capacitors, the energy stored in the 6 pF capacitor is E. Find the value of the following :

1. Energy is stored in a 12 μF capacitor.
2. Energy is stored in 3 μF capacitor.
3. Total energy drawn from the battery

Answer:

1. Let the EMF of the applied battery be V.

Energy stored in 6 μF capacitor = E (given)

Energy stored in a capacitor is given by

U = \(\frac{1}{2} \mathrm{CV}^2 \Rightarrow \mathrm{V}=\sqrt{\frac{2 \mathrm{U}}{\mathrm{C}}}\)

Here \(V_1=V_2\)

⇒ \(V_1 \Rightarrow \text { Potential difference across } 6 \mu \mathrm{F} \text { capacitor }\)

⇒ \(V_2 \Rightarrow \text { Potential difference across } 12 \mu \mathrm{F} \text { capacitor }\)

⇒ \(\sqrt{\frac{2 \mathrm{E}_1}{\mathrm{C}_1}}=\sqrt{\frac{2 \mathrm{E}_2}{\mathrm{C}_2}}\)

⇒ \(\mathrm{E}_1=\mathrm{E} \text { (given) }\)

⇒ \(\mathrm{E}_2=\frac{\mathrm{C}_2}{\mathrm{C}_1} \times \mathrm{E}_1 \Rightarrow \mathrm{E}_2=\frac{12}{6} \times \mathrm{E} \Rightarrow \mathrm{E}_2=2 \mathrm{E}\)

2. Energy stored in 12 μF capacitor (E₂) = 2E

The total energy of a parallel combination of capacitors is the sum of the energy of both capacitors.

⇒ E + 2E = 3E

So, \(\frac{Q^2}{2 C_{c i}}=3 E \Rightarrow Q=\sqrt{2 C_{c i} \times 3 E}\)

The same charge (Q) flows through 3μF capacitor. So energy stored by a capacitor of 3 μF

⇒ \(\mathrm{E}_3=\frac{\mathrm{Q}^2}{2 \mathrm{C}_3}=\frac{2 \mathrm{C}_{\mathrm{ct} .} \times 3 \mathrm{E}}{2 \mathrm{C}_3} \Rightarrow \mathrm{E}_3=\frac{2 \times 18 \times 3 \mathrm{E}}{3 \times 2}\)

⇒ \(\mathrm{E}_3=18 \mathrm{E}\)

Total energy stored \(=\mathrm{E}_1+\mathrm{E}_2+\mathrm{E}_3\)

= E + 2E+ 18 E = 21 E

Thus, the energy is given by the battery = 2 x 21E = 42E

[As half part of the total energy given by the battery is stored in capacitors]

Question 12. Find the ratio of the potential differences that must be applied across the series and parallel combination of two capacitors C₁ and C₂ with their capacitances in the ratio 1: 2 so that the energy stored in the two eases becomes the same
Answer:

Let C₁ =x and C₂ = 2x

Equivalent capacitance in series combination \(C_{\mathrm{S}}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{\mathrm{x} \times 2 \mathrm{x}}{\mathrm{x}+2 \mathrm{x}}=\frac{2 \mathrm{x}}{3}\)

∴ \(\mathrm{C}_{\mathrm{s}}=\frac{2 \mathrm{x}}{3}\)

Equivalent capacitance in parallel combination

C_P = C₁ + C₂ = x + 2x = 3x

Now given that energy stored in series combination = Energy stored in parallel combination

⇒ \(\frac{1}{2} C_s V_1^2=\frac{1}{2} C_p V_2^2 \Rightarrow \frac{1}{2} \times\left(\frac{2 x}{3}\right) V_1^2=\frac{1}{2} \times 3 x \times V_2^2 \Rightarrow \frac{V_1}{V_2}=\frac{3}{\sqrt{2}}\)

Question 13. Write the definition of electric potential. Calculate the electric potential due to a point charge Q at a distance r from it. Draw a graph between electric potential V and distance r for a point charge Q.
Answer:

Electric potential: The amount of work needed to bring a unit positive charge from infinity to a specific point inside the electric field of a positive charge is called electric potential at that point.

Potential due to point charge: Electrostatic force on + q₀ at point A due to charge Q

⇒ \(\mathrm{F}_{\mathrm{c}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q} \mathrm{q}_0}{\mathrm{x}^2}\)

Work done in moving a charge +q₀ in short displacement from A to B

dW = \(\mathrm{F}_c \mathrm{dx}=-\frac{\mathrm{Qq}_0}{4 \pi \varepsilon_0} \times \frac{1}{\mathrm{x}^2} \mathrm{dx}\)

Total work done in moving a charge q₀ from ∞ to r

W = \(-\frac{\mathrm{Qq}_0}{4 \pi \varepsilon_0} \int_0^{\mathrm{r}} \frac{1}{\mathrm{x}^2} \mathrm{dx}\)

W = \(-\frac{\mathrm{Qq_{0 }}}{4 \pi \varepsilon_0}\left[-\frac{1}{\mathrm{x}}\right]^{\mathrm{r}}_{\mathrm{∞}}\)

W = \(\frac{\mathrm{Qq_{0 }}}{4 \pi \varepsilon_0}\left[\frac{1}{\mathrm{r}}-\frac{1}{\infty}\right]^{\mathrm{r}}_{\mathrm{∞}} \Rightarrow \mathrm{W}=\frac{\mathrm{Qq_{0 }}}{4 \pi \varepsilon_{0} \mathrm{r}}\)

From the definition of potential

V = \(\frac{W}{q_0} \Rightarrow V=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r}\)

V = \(\frac{k Q}{r} \Rightarrow V \propto \frac{1}{r}\)

Electrostatic Potential And Capacitance Long Answer Type Questions

Question 1. A dielectric slab is a substance that does not allow the flow of charges through it but permits it to exert electrostatic forces on one another. When a dielectric slab is placed between the plates, the field E₀ polarises the dielectric. This induces charge -Q_P on the upper surface and +Q_P on the lower surface of the dielectric. These induced charges set up a field E_P inside the dielectric in the opposite direction of E₀ as shown.

1. In a parallel plate capacitor, the capacitance increases from 4μF to 80μF upon introducing a dielectric medium what is the dielectric constant of the medium?

1. 10
2. 20
3. 50
4. 100

Answer: 2. 20

2. A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the plates is now reduced by half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in the second case.

1. 8pF
2. 10pF
3. 80pF
4. 100pF

Answer: 3. 80pF

3. A dielectric introduced between the plates of a parallel plate capacitor :

1. Increases the electric field between the plates
2. Decreases the capacity of the capacitor
3. Increases the charge stored in the capacitor
4. Increases the capacity of the capacitor

Answer: 4. Increases the capacity of the capacitor

4. A parallel plate capacitor of capacitance 1 pF has separation between the plates is d. When the distance of separation becomes 2d and wax of dielectric constant x is inserted in it the capacitance becomes 2 pF. What is the value of x?

1. 2
2. 4
3. 6
4. 8

Answer: 2. 4

Quetsion 2. The electrostatic potential of a charged body represents the degree of electrification of the body. It determines the flow of electric charge between two charged bodies placed in contact with each other. The charge always flows from a body of higher potential to another body at lower potential. The flow of charge stops as soon as the potential of the two bodies becomes equal. Now answer the following.

1. A uniform electric field of 100 N/C exists vertically upward direction. The decrease in electric potential as one goes up through a height of 5cm is :

1. 20V
2. 120V
3. 5V
4. Zero

Answer: 3. 5V

2. Work done to bring a unit positive charge unaccelerated from infinity to a point inside the electric field is called

1. Electric Field
2. Electric Potential
3. Capacitance
4. Electric Flux

Answer: 2. Electric Potential

3. Electric potential at a distance of 27 cm from a point charge 9 NC, is given by

1. 200 V
2. 100 V
3. 300 V
4. 50V

Answer: 3. 300 V

4. Two charges -6μC and 10μC arc placed at a distance 18 cm apart. The electric potential at the midpoint joining these two will be

1. 4 x 10⁵ V
2. 4x 10⁴ V
3. 4x 10³ V
4. 4x 10⁶ V

Answer: 1. 4 x 10⁵ V

Question 2.

1. Derivr an expression for the capacitance of a parallel plate capacitor filled with a medium of dielectric constant K.
2. A charge q = 2 μC is placed at the center of a sphere of radius 20 cm. What is the amount of work done in moving 4 μC from one point to another point on its surface?
3. Write a relation for polarisation \(\vec{P}\) of a dielectric material in the presence of an external electric field.

Answer:

Electrical capacitance: The ability of a conductor to store electrical energy (or charge) is called electric capacitance. The charge which is given to a conductor is directly proportional to its increasing potential

q ∝ V ⇒ q=C V

⇒ C = \(\frac{q}{V}\)

C ⇒ electrical capacity of a conductor.

The capacitance of parallel plate capacitor :

Electric Held intensity between two plates of the capacitor.

⇒ \(E_m=\frac{\sigma}{\varepsilon}\left\{\begin{array}{l}
\sigma=\frac{q}{A} \\
\varepsilon=\varepsilon_0 K
\end{array} \Rightarrow E_m=\frac{q}{\varepsilon_0 K A}\right.\)

The potential difference between the two plates of the capacitor

V = \(E_m \times d\)

V = \(\frac{q d}{\varepsilon_0 K A}\)

Capacitance C = \(\frac{q}{V} \Rightarrow C=\frac{q}{\left(\begin{array}{c}q d \\ \varepsilon_0 K A\end{array}\right)} \Rightarrow C=\frac{\varepsilon_0 K A}{d}\)

2. Sphere is equipotential surface: So. work done will be zero.

3. \(\vec{p}\)= ε₀χ₀\(\vec{E}\)

Question 3.

1. Distinguish, with the help of a suitable diagram, the difference in the behavior of a conductor and a dielectric placed in an external electric field. Mow docs polarised dielectric modify the original external field.
2. A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?
   1. The charge is stored by the capacitor
   2. field strength between the plates.
   3. Energy stored by the capacitor
   4. Justify your answer with ease.

Answer:

1. When a conductor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that the electric field due to induced charges opposes the external field within the conductor. This happens until a sialic situation is achieved, i.e. when the two fields cancel each other and the net electrostatic field in the conductor becomes zero.

In contrast to conductors, dielectrics are non-conducting substances, i.e. they have no charge carriers. Thus, in a dielectric, free movement of charges is not possible. It turns out that the external field induces dipole moment by stretching molecules of the dielectric.

The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produces a field that opposes the external field. However, the opposing field is so induced, that does not exactly cancel the external field. It only reduces it. The extent of the effect depends on the nature of the dielectric.

Both polar and non-polar dielectrics develop net dipole moments in the presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by P for linear isotropic dielectrics

2. After disconnecting the battery and doubling the separation between the two plates of the capacitor

(1) Charge on capacitor remains same (due to charge conservation)

i.e., \(C V=C^{\prime} V^{\prime} \Rightarrow C V=\left(\frac{C}{2}\right) V^{\prime} \Rightarrow V^{\prime}=2 V\)

Voltage will be doubled

(2) Field strength between the plates.

E’ = \(\frac{V^{\prime}}{d^{\prime}}=\frac{2 V}{2 d} \Rightarrow E^{\prime}=\frac{V}{d}=E\)

The electric field between the plates remains the same.

(3) Energy stored in capacitor when connected to battery \(\mathrm{U}_1=\mathrm{q}^2 / 2 \mathrm{C}\)

Now energy is stored in the capacitor after disconnection from the battery

⇒ \(U_2=\frac{q^2}{2 C^{\prime}}=\frac{q^2}{2 \times C / 2}=q^2 / C\)

∴ Energy stored in the capacitor gets doubled to its initial value.

Question 4.

1. Two parallel plate capacitors X and Y have the same area of plates and the same separation between them. X has air between the plates while Y contains a dielectric medium of ε_r=4.
   1. Calculate the capacitance of each capacitor if the equivalent capacitance of the combination is 4 μF
   2. Calculate the potential difference between the plates of X and Y.
   3. Estimate the ratio of electrostatic energy stored in X and Y.
2. Two metallic spheres of radii R and 2R are charged so that both of these have the same surface charge density σ. If they are connected to each other with a conducting wire, in which direction will the charge flow and why?

Answer:

1. (1) Let us assume that capacitance of X capacitor = C μF

∴ Capacitance of Y capacitor = 4C μF (As C = KC) equivalent capacitance of system = 4μF (given)

So, \(\frac{1}{4 C}+\frac{1}{C}=\frac{1}{4} \Rightarrow C=5 \mu \mathrm{F}\)

The capacitance of Y capacitor → 4C = 20pF

(2) We assume the potential of capacitor X is V₁ and V₂ for capacitor Y

∴ \(\mathrm{Q}_{\mathrm{X}}=\mathrm{Q}_{\mathrm{Y}}\) [Both capacitor are in series]

⇒ \(C_X V_X=C_Y V_Y\)

⇒ \(\frac{1}{4}=\frac{V_Y}{V_X} \Rightarrow V_Y=\frac{V_X}{4}\)….(1)

Also given that Vx + VY = 15V…(2)

By solving the above eq (1) and (2),

we get V₁ = 12V V_Y = 3V

(3) Ratio of energy stored \(\frac{E_X}{E_Y}=\frac{C_X V_X^2}{C_Y V_Y^2}=\frac{C_X}{C_Y} \times\left(\frac{V_X}{V_Y}\right)^2=\frac{1}{4} \times\left(\frac{4}{1}\right)^2 \Rightarrow \frac{E_X}{E_Y}=\frac{4}{1}\)

2. Potential on the surface of a charged metallic sphere is given by

V = kq/r

V= k.4π² σ/r

So V ∝ r

It means that a sphere having a large radius will be at a higher potential and charge always flows from a higher potential to a lower potential. So charge will flow from the metallic sphere of radius 2R to the sphere of radius R.

 

 

 

 

 

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