## Capacitance And Capacitor Multiple Choice Question And Answers

**Question 1. A capacitor of 4μF connected as shown in the circuit. The internal resistance of the battery is 0.5 H. The amount of charge on the capacitor plates will be**

- 0
- 4μC
- 16μC
- 8μC

**Answer:** 4. 8μC

⇒ \(I=\frac{E}{R}=\frac{2.5}{2+0.5}=1 {\mathrm{A}}\)

V= E-IR

= 2.5- 1 x 10.5

= 2 V

∴ Q = CV

= 4 x 2μC

= 8μC

**Question 2. In the circuit, initially, key K _{1} is closed and K_{2} is open. Then K_{1} is opened and K_{2} is closed(order is important). Take Q_{1} and Q_{2} as charges on C_{1} and C_{2} and V_{1} and V_{2} as voltage respectively. Then**

**Read and Learn More Class 12 Physics Multiple Choice Questions**

- Charge on C
_{1}gets redistributed such that V_{1}= V_{2} - Charge on C
_{1}gets redistributed such that Q’_{1}= Q_{2} - Charge on C
_{1}gets redistributed such that C_{1}V_{1}+ C_{2}V_{2}= C_{1}E - Charge on C
_{1}gets redistributed such that Q’_{2}+Q’_{2}= Q

**Answer:**

1. Charge on C_{1} gets redistributed such that V_{1} = V_{2}

4. Charge on C_{1} gets redistributed such that

When K_{1} is closed and K_{2} is open, Q (on C_{1}) =EC_{1}.

When K_{2} is closed and K_{1} is open, there being no battery in the circuit, C_{1} and C_{2} are connected in parallel and the charge on C_{1} will be redistributed between C_{1} and C_{2} in such a manner that V_{1} = V_{2}.

Since there is no loss of charge, Q = Q’_{1} + Q’_{2 }

**Question 3. A parallel plate capacitor is connected to a battery.**

**Consider two situations:**

**A. Key K is kept dosed and plates of capacitors are moved apart using insulating handles.**

**B. Key A’ is opened and plates of capacitors are moved apart using insulating handles.**

**Choose the correct option(s).**

**In A:**Q remains the same but C changes**In B:**V remains the same but C changes**In A:**V remains the same and hence Q changes**In B:**Q remains the same ami hence V changes

**Answer:**

**3. In A:** V remains the same and hence Q changes

**4. In B:** Q remains the same ami hence V changes

⇒ \(Q=C E=\frac{\epsilon_0 A}{d} \cdot B\)

Thus if d increases, Q will decrease.

⇒ \(V=\frac{Q}{C}=\frac{Q d}{\epsilon_0 A}\)

∴ If d decreases, Q remains the same, and V increases.

**Question 4. A solid sphere and a hollow sphere of the same diameter are charged to the same potential. Then**

- The charge on the hollow sphere will be greater
- Both spheres will have the same charge
- Only the hollow sphere will be charged
- The solid sphere will have a greater amount of charge

**Answer:** 2. Both the spheres will have the same charge

**Question 5. 64 water drops coalesce to form a big drop. If each small drop has capacity C, potential V and charge Q, the capacitance of the big drop will be**

- C
- 4C
- 16C
- 64C

**Answer:** 2. 4C

**Question 6. If the radius of a conducting sphere is 1m, its capacitance in farad will be**

- 10
^{-3} - 10
^{-6} - 9 x 10
^{-9} - 1.1 x 10
^{-10}

**Answer:** 4. 1.1 x 10^{-10}

**Question 7. n small drops of the same size are charged to Vvolt each. They coalesce to form a big drop. The potential of a big drop will be**

- n
^{1/3}V - n
^{2/3}V - n
^{3/2}V - n
^{3}V

**Answer:** 2. n^{2/3}V

**Question 8. A conducting sphere of radius 10 cm is kept in a medium of dielectric constant 8. Its capacitance is**

- 80 esu
- 10 esu
- \(\frac{1}{9}\) x 10
^{-10}F - 80F

**Answer:** 1. 80 esu

**Question 9. When two charged conductors are joined by a thin conducting wire, charge flows from one to the other, until**

- Charges on them become equal
- Their capacitances become equal
- Their potential becomes equal
- Energy stored in them becomes equal

**Answer:** 3. Their potentials become equal

**Question 10. Two conducting spheres of radii r _{1} and r_{2} are connected by a thin conducting wire. Some amount of charge is given to the system. The charge will be shared between them in such a way that the ratio between the surface densities of charge on the spheres is equal to**

- \(\frac{r_1}{r_2}\)
- \(\frac{r_2}{r_1}\)
- \(\frac{r_1^2}{r_2^2}\)
- \(\frac{r_2^2}{r_1^2}\)

**Answer:** 2. \(\frac{r_2}{r_1}\)

**Question 11. Two capacitors C _{1} and C_{2} (= 2C_{1}) are connected in a circle with a switch. Initially, the switch is open and C_{1} holds charge Q. The switch is closed. At a steady state, the charge on each capacitor would be**

- Q, 2Q
- \(\frac{Q}{3}, \frac{2 Q}{3}\)
- \(\frac{3 Q}{2}, 3 Q\)
- \(\frac{2 Q}{3}, \frac{4 Q}{3}\)

**Answer:** 2. \(\frac{Q}{3}, \frac{2 Q}{3}\)

**Question 12. When a capacitor is connected to a dc battery,**

- No current flows through the circuit
- Current flows through the circuit for some time, but eventually stops
- The current grows up and reaches a maximum value when the capacitor is fully charged
- The current reverses its direction alternately due to the charging and discharging of the capacitor

**Answer:** 2. Current flows through the circuit for some time, but eventually stops

**Question 13. A few capacitors are equally charged. Which of the nature of variation of the potential difference V between their plates with their capacitances C?**

**Answer:** 4.

**Question 14. A parallel plate capacitor is charged and then disconnects the battery. If the plates of the capacitor are moved farther apart—**

- The charge on the capacitor will increase
- The potential difference between the two plates will increase
- The capacitance will increase
- The electrostatic energy stored in the capacitor will decrease

**Answer:** 2. The potential difference between the two plates will increase

**Question 15. The capacity of a parallel plate capacitor is proportional to the power n of the distance between the plates. The value of n is**

- 1
- 2
- -1
- -2

**Answer:** 3. -1

**Question 16. The capacitance of a parallel plate capacitor depends on**

- The thickness of the plates
- The charge accumulated on the plates
- The potential difference between the two plates
- The distance between the two plates

**Answer:** 4. The distance between the two plates

**Question 17. The separation between the two plates of a parallel plate ****capacitor is d. A metal plate of equal area and of thickness ****\(\frac{d}{2}\) is inserted between the two plates. The ratio of the capacitances after and before this insertion is**

- 2: 1
- A√2:1
- 1:72
- 1:2

**Answer:** 1. 2: 1

**Question 18. A parallel plate air capacitor is connected to a battery and is then disconnected. Now the intermediate space is filled up with a medium of specific inductive capacity K. The potential difference between the plates will change by a factor of**

- k
_{2} - k
- \(\frac{1}{k}\)
- \(\frac{1}{\kappa^2}\)

**Answer:** 3. \(\frac{1}{k}\)

**Question 19. A parallel plate capacitor with oil (dielectric constant K = 2) has a capacitance C. If the oil is removed, then the capacity of the capacitor becomes**

- √2C
- 2C
- \(\frac{C}{\sqrt{2}}\)
- \(\frac{C}{2}\)

**Answer:** 4. \(\frac{C}{2}\)

**Question 20. A parallel plate capacitor has a capacitance of 100μF. The plates are at a distance d apart. A slab of thickness t(t<d) and dielectric constant 5 is introduced between the parallel plates. Then capacitance can be,**

- 50μF
- 100μF
- 200μF
- 500μF

**Answer:** 3. 200μF

**Question 21. Two identical capacitors of C _{1} and C_{2} are connected to a battery. C_{1} is filled with air and C_{2} is filled with an insulator of dielectric constant ∈r then**

- Q
_{1}> Q_{2} - Q
_{1}< Q_{2} - Q
_{1}= Q_{2} - None

**Answer:** 2. Q_{1} < Q_{2}

**Question 22. A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now**

- V
- V + \(\frac{Q}{C}\)
- F + \(\frac{Q}{2C}\)
- None

**Question 23. A combination is formed by connecting alternately n number of equidistant parallel plates. If C be the capacitance for any two consecutive plates, then the capacitance of the whole system will be**

- C
- nC
- (n-1)C
- (n + 1)C

**Answer:** 3. (n-1)C

**Question 24. Two capacitors of capacitances C _{1} and C_{2}, are connected in parallel. A charge q given to the combination is distributed between the two. The ratio between these charges on the two capacitors is**

- \(\frac{C_1}{C_2}\)
- \(\frac{C_2}{C_1}\)
- \(\frac{C_1 C_2}{1}\)
- \(\frac{1}{C_1 C_2}\)

**Answer:** 1. \(\frac{C_1}{C_2}\)

**Question 25. The equivalent capacitance of the combination is shown between A and B.**

- \(\frac{C}{3}\)
- \(\frac{C}{2}\)
- \(\frac{2}{3}\)C
- C

**Answer:** 1. \(\frac{C}{3}\)

**Question 26. Each of the four horizontal metal plates has a surface area a, and the separation between each pair of consecutive plates is d. The plates are connected to points A and B. The equivalent capacitance between A and B, when the system is kept in air, is**

- \(\frac{\epsilon_0 \alpha}{d}\)
- \(\frac{2 \epsilon_0 \alpha}{d}\)
- \(\frac{3 \epsilon_0 \alpha}{d}\)
- \(\frac{4 \epsilon_0 \alpha}{d}\)

**Answer:** 2. \(\frac{2 \epsilon_0 \alpha}{d}\)

**Question 27. In the connection the equivalent capacitance between A and B is**

- 8μF
- 4μF
- 2μF
- 3μF

**Answer:** 2. 4μF

**Question 28. The charge on any of the two 2μF capacitors and that on the 1μF capacitor, in μC unit, are**

- 1 and 2
- 2 and 1
- 1 and 1
- 2 and 2

**Question 29. The equivalent capacitance is C _{1} when four capacitors of equal capacitance are connected in series. In their parallel connection, the equivalent capacitance becomes C_{2}. The ratio C_{1}/C_{2} will be**

- \(\frac{1}{4}\)
- \(\frac{1}{16}\)
- \(\frac{1}{8}\)
- \(\frac{1}{12}\)

**Answer:** 2. \(\frac{1}{16}\)

**Question 30. Three plates, each of area A, are connected. The effective capacitance will be**

- \(\frac{\epsilon_0 A}{d}\)
- \(\frac{3 \epsilon_0 A}{d}\)
- \(\frac{3}{2} \frac{\epsilon_0 A}{d}\)
- \(\frac{2 \epsilon_0 A}{d}\)

**Answer:** 4.

**Question 31. In the area of each of the four plates P, Q, R, and S is a, and the separation between consecutive plates is d. The equivalent capacitance between A and B is**

- \(\frac{4 \epsilon_0 \alpha}{d}\)
- \(\frac{3 \epsilon_0 \alpha}{d}\)
- \(\frac{2 \epsilon_0 \alpha}{d}\)
- \(\frac{\epsilon_0 \alpha}{d}\)

**Answer:** 2. \(\frac{3 \epsilon_0 \alpha}{d}\)

**Question 32. Five capacitors, each of capacitance C, are connected. The ratio of the capacitance between P and R to that between P and Q is**

- 3:1
- 5: 2
- 2: 3
- 1: 1

**Answer:** 3. 2: 3

**Question 33. There are six capacitors in the given network. The capacitance of each is 4μF.**

**The effective capacitance between A and B is**

- 24μF
- 12μF
- \(\frac{3}{2}\)μF
- \(\frac{2}{3}\)μF

**Answer:** 1. 24μF

**Question 34. Find the equivalent capacitance between A and B.**

- 5μF
- 4μF
- 3μF
- 2μF

**Answer:** 1. 5μF

**Question 35. A gang condenser is formed by interlocking a number of plates. The distance between the consecutive plates is 0.885 cm and the overlapping area of the plates is 5 cm². The capacity of the unit is**

- 1.06 pF
- 4pF
- 6.36 pF
- 12.72 pF

**Question 36. A number of condensers of equal capacity C are arranged in n columns. The equivalent capacity is given by**

- nC
- n(n + 1)C
- \(\frac{n(n+1)}{2} C\)
- 2n(n + 1)C

**Question 37. Three capacitors connected in series have an effective capacitance of 2μF. If one of the capacitors is removed, the effective capacitance becomes 3μF. The capacitance of the capacitor that is removed is**

- 1μF
- \(\frac{3}{2}\)μF
- \(\frac{2}{3}\)μF
- 6μF

**Answer:** 4. 6μF

**Question 38. In a charged capacitor, energy is**

- Equally shared between the positive and the negative plates
- Stored in one plate when the other is grounded
- Stored in the electric field between the two plates
- Discharged if one of the plates is grounded

**Answer:** 3. Stored in the electric field between the two plates

**Question 39. A parallel plate capacitor, with a slab of dielectric constant K between its plates, has a capacitance C. It is charged to a potential V. Now the dielectric slab is first brought out of the capacitor and is then introduced again. The work done in this process will be**

- \((\kappa-1) \frac{C V^2}{2}\)
- \(\frac{C V^2(\kappa-1)}{\kappa}\)
- (k – 1)CV
^{2} - 0

**Answer:** 4. 0

**Question 40. The work done to charge a capacitor of capacitance 100μF with 8 x 10 ^{-18}C will be**

- 32 X 10
^{-32}J - 16 X 10
^{-32}J - 3.2 X 10
^{-26}J - 4 X 10
^{-10}J

**Answer:** 1. 32 X 10^{-32}J

**Question 41. A battery continues to charge a parallel plate capacitor until the potential difference between its plates becomes equal to the emf of the battery. The ratio of the energy stored in the tire capacitor to tire work done by the battery will be**

- 1
- \(\frac{2}{1}\)
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)

**Answer:** 3. \(\frac{1}{2}\)

**Question 42. The space between the plates of a parallel plate air capacitor is filled with a material of dielectric constant K, keeping the plates connected to a certain external battery. The energy stored in this capacitor will change by a factor of**

- k
^{2} - k
- \(\frac{1}{k}\)
- \(\frac{1}{\kappa^2}\)

**Answer:** 2. k

**Question 43. A parallel plate capacitor is connected to a battery. The plates are pulled apart at a uniform speed. If x is the separation between the plates, the time rate of change of electrostatic energy of the capacitor is proportional to**

- x
^{-2} - x
^{-2} - x
^{-1} - x
^{2}

**Answer:** 1. x^{-2}

**Question 44. If the potential difference between the plates of a capacitor is increased by 20%, the energy stored in the capacitor increases by exactly**

- 20%
- 22%
- 40%
- 44%

**Answer:** 4. 44%

**Question 45. If the distance between the plates of a parallel plate capacity is doubled while the battery is kept connected, then**

- The charge stored in the capacitor will be doubled
- The battery will absorb some amount of energy
- The electric field between the two plates will be halved
- Work will be done by an external agent on the two plates

**Answer:** 3. The electric field between the two plates will be halved

**Question 46. Which of the following gas can be used in V a de Graaff generator?**

- Methane
- Hydrogen
- Oxygen
- Chlorine

**Answer:** 1. Methane

**Question 47. Van de graaff generator is used as**

- External voltage source
- External current source
- Electrostatic accelarator
- None of these

**Answer:** 3. Electrostatic accelerator

**Question 48. A parallel plate capacitor is charged and the charging battery is disconnected. If the plates of the capacitor are moved further apart by means of insulating handles. Then**

- The charge on the capacitor increases
- The voltage across the plates increases
- The capacitance of the capacitor increases
- The electrostatic energy started in the capacitor increases

**Answer:**

2. The voltage across the plates increases

4. The electrostatic energy started in the capacitor increases

**Question 49. The same potential difference is applied between A and B. If C is joined to D,**

- No charge will flow between C and D
- Some charge will flow between C and D
- The equivalent capacitance between C and D will not change
- The equivalent capacitance between C and D will change

**Answer:**

1. No charge will flow between C and D

3. The equivalent capacitance between C and D will not change

**Question 50. A parallel plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E, and W denote respectively the charge on each plate, the electric field between the plates (after the slab is inserted), and the work done in the process of inserting the slab, then**

- \(Q=\frac{\epsilon_0 A V}{d}\)
- \(Q=\frac{\epsilon_0 K A V}{d}\)
- \(E=\frac{V}{\kappa d}\)
- \(W=\frac{\epsilon_0 A V^2}{2 d}\left(1-\frac{1}{k}\right)\)

**Answer:**

1. \(Q=\frac{\epsilon_0 A V}{d}\)

3. \(E=\frac{V}{\kappa d}\)

4. \(W=\frac{\epsilon_0 A V^2}{2 d}\left(1-\frac{1}{k}\right)\)

**Question 51. A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 3d. The slab is equidistant from the plates. The capacitor is given some charge by connecting it to a battery. As x goes from 0 to 3d,**

- The magnitude of the electric field remains the same
- The direction of the electric field remains the same
- The electric potential increases continuously
- The electric potential increases at first, then decreases, and again increases

**Answer:**

2. The direction of the electric field remains the same

3. The electric potential increases continuously

**Question 52. In the circuit, the potential difference across the 3μF capacitor is V, and the equivalent capacitance between A and B is CAB. Correct relations are**

- C
_{AB}= 4μF - \(C_{A B}=\frac{18}{11} \mu \mathrm{F}\)
- V = 20 V
- V = 40 V

**Answer:**

1. C_{AB }= 4μF

4. V = 40 V

**Question 53. In the given circuit C _{1} = C_{2} = 2μF. Then charge is stored in**

- Capacitor C
_{1}is zero - Capacitor C
_{2}is zero - Both C
_{1}and C_{2}is zero - Capacitor C
_{1}is 40μC

**Answer:**

2. Capacitor C_{2} is zero

4. Capacitor C_{1} is 40μC

**Question 54. In the given network which one is correct?**

- \(\left|q_2\right|=280 \mu \mathrm{C}\)
- \(\left|q_3\right|=160 \mu \mathrm{C}\)
- q
_{2}= 120μC, q_{3}= 0 - Impossible to find q
_{2}and q_{3}unless C_{1}and Q_{2}known

**Answer:**

1. \(\left|q_2\right|=280 \mu \mathrm{C}\)

2. \(\left|q_3\right|=160 \mu \mathrm{C}\)

**Question 55. A parallel plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor. Then**

- The battery will supply the same charge
- The capacitance will increase
- The potential difference between the plates will increase
- Equal and opposite charges will appear on the two faces of the metal plate

**Answer:**

2. The capacitance will increase

4. Equal and opposite charges will appear on the two faces of the metal plate

**Question 56. A capacitor of capacitance C _{1} is charged up to potential V and then connected in parallel to an uncharged capacitor capacitance C_{2}. The final potential difference across each capacitor will be**

- \(\frac{C_2 V}{C_1+C_2}\)
- \(\frac{C_1 V}{C_1+C_2}\)
- \(\left(1+\frac{C_2}{C_1}\right) V\)
- \(\left(1-\frac{C_2}{C_1}\right) V\)

**Answer:** 2. \(\frac{C_1 V}{C_1+C_2}\)

The charge storedin the first capacitor, Q = C_{1} V When the charged capacitor C_{1} is connected in parallel to the uncharged capacitor C_{2}, then the equivalent capacitance of the combination, C_{eq} = C_{1} + C_{2}. If Vf is the final potential difference across each capacitor,

then \(V_f=\frac{Q}{C_{\mathrm{eq}}}=\frac{C_1 V}{C_1+C_2}\)

The option 2 is correct.

**Question 57. 64 small water drops each of capacitance C and charge q coalesce to form a larger spherical drop. The charge and capacitance of the larger drop is**

- 64q, C
- 16q, 4C
- 64q, 4C
- 16q, C

**Answer:** 3. 64q, 4C

Total charge = 64q

If the radius of each small water drop is r and that of the large water drop is R,

⇒ \(\frac{4}{3} \pi R^3=64 \times \frac{4}{3} \pi r^3\)

or, R³ = 64r³

or, R = 4r

The capacitance of a spherical drop or radius of the drop.

So,

capacitance of the large water drop = C x \(\frac{R}{r}\) = 4C

The option 3 is correct.

**Question 58. Three capacitors, 3μF, 6μF, and 6μF series to a source of 120 V. The potential difference, in volts, across the 3μF capacitor will be**

- 24
- 30
- 40
- 60

**Answer:** 4. 60

Equivalent capacitance of the combination,

⇒ \(C_{\mathrm{eq}}=\frac{1}{\frac{1}{3}+\frac{1}{6}+\frac{1}{6}}\)

= \(\frac{3}{2} \mu \mathrm{F}\)

= \(\frac{3}{2} \times 10^{-6} \mathrm{~F}\)

∴ The total charge of the combination,

⇒ \(Q=C_{\mathrm{eq}} \cdot V\)

= \(\frac{3}{2} \times 10^{-6} \times 120 \mathrm{C}\)

= \(180 \times 10^{-6} \mathrm{C}\)

∴ The potential difference across the capacitor of capacitance 3μF

⇒ \(\frac{180 \times 10^{-6}}{3 \times 10^{-6}}=60 \mathrm{~V}\)

The option 4 is correct.

**Question 59. Consider two concentric spherical metal shells of radii r _{1} and r_{2} (r_{2} > r_{1}). If the outer shell has a charge q and the inner one is grounded, the charge on the inner shell is**

- \(-\frac{r_2}{r_1} q\)
- 0
- \(-\frac{r_1}{r_2} q\)
- -q

**Answer:** 3.

As the inner sphere is connected to the ground, its potential is zero.

∴ \(\frac{1}{4 \pi \epsilon} \frac{q_1}{r_1}+\frac{1}{4 \pi \epsilon} \frac{q_2}{r_2}=0\) [q1 = charge of the inner sphere, q2 = charge of the outer sphere]

i.e., \(q_1=-\left(\frac{r_1}{r_2}\right) q_2\)

The option 3 is correct.

**Question 60. Half of the space between the plates; of a parallel plate capacitor is filled with a dielectric material of dielectric constant Kl The remaining contains air. The capacitor is now given a charge Q. Then**

- The electric field in the dielectric region is higher than that in the air-filled region
- On the two halves of the bottom plate, the charge densities are unequal
- Charge on the half of the top plate above the air-filled part is \(\frac{Q}{k+1}\)
- The capacitance of the capacitor shown above is \(\frac{(1+\kappa) C_0}{2}\) where C0 is the capacitance of the same capacitor with the dielectric removed

**Answer:**

1. The electric field in the dielectric region is higher than that in the air-filled region

2. On the two halves of the bottom plate, the charge densities are unequal

3. Charge on the half of the top plate above the air-filled part is \(\frac{Q}{k+1}\)

4. The capacitance of the capacitor shown above is \(\frac{(1+\kappa) C_0}{2}\) where C0 is the capacitance of the same capacitor with the dielectric removed

⇒ \(C_0=\frac{\epsilon_0 A}{d}\)

∴ The capacitance of the capacitor

**⇒ **\(C=\frac{\kappa \epsilon_0 \frac{A}{2}}{d}+\frac{\epsilon_0 \frac{A}{2}}{d}\)

= \(\frac{\epsilon_0 \frac{A}{2}}{d}(\kappa+1)\)

= \((1+\kappa) \frac{C_0}{2}\)

The option 4 is correct.

∴ \(V=\frac{\sigma_1}{\kappa \epsilon_0}=\frac{\sigma_2}{\epsilon_0}\) [Here dielectric and air medium is denoted by the

subscripts 1 and 2respectively.]

or, σ_{1} = k σ_{2}

∴ σ_{1} ≠ σ_{2}

The option 2 is correct.

Now, \(Q_1=\frac{\sigma_1 A}{2} ; Q_2=\frac{\sigma_2 A}{2}\)

∴ \(\frac{Q_1}{Q_2}=\frac{\sigma_1}{\sigma_2}=\frac{1}{\kappa}\)

or, \(\frac{Q_1}{Q}=\frac{Q_2}{Q_1+Q_2}=\frac{1}{\kappa+1}\)

or, \(Q_2=\frac{Q}{\kappa+1}\)

The option 3 is correct.

⇒ \(E_1=\frac{V}{d}=E_2\)

The option 1 is not correct.

**Question 61. A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then**

- The Energy Stored In The Capacitor Decreases
- The Capacitance Of The Capacitor Increases
- The Charge In The Capacitor Decreases
- The voltage across the capacitor increases

**Answer:** 4. The voltage across the capacitor increases

The potential difference between the two plates,

⇒ \(V=\text { electric field }(E) \times \text { distance }(d)=\frac{\sigma}{\kappa \epsilon_0} \times d\)

∴ V ∝ d

i.e., the potential difference across the capacitor increases with an increase in distance between the plates.

The option 4 is correct.

**Question 62. A 5μF capacitor is connected in series with a 10μF capacitor. When a 300-volt potential difference is applied across this combination, the total energy stored in the capacitors is**

- 15 J
- 1.5 J
- 0.15 J
- 0.10 J

**Answer:** 3. 0.15 J

Equivalent capacitance,

⇒ \(C=\frac{5 \times 10}{5+10}\)

= \(\frac{50}{15}=\frac{10}{3} \mu \mathrm{F}\)

= \(\frac{10}{3} \times 10^{-6} \mathrm{~F}\)

= \(\frac{1}{3} \times 10^{-5}\)

∴ Total energy stored = \(\frac{1}{2} C V^2=\frac{1}{2} \times\left(\frac{1}{3} \times 10^{-5}\right) \times 300^2\)

= 0.15 J

The option 3 is correct

**Question 63. Equivalent capacitance between A and E**

- 20μF
- 8μF
- 12μF
- 16μF

**Answer:** 2. 8μF

The equivalent circuit

Hence, the equivalent capacitance between A and B

C_{eq} = 2 + 4 + 2

= 8μF

The option 2 is correct.

**Question 64. A 1μF capacitor C is connected to a battery of 10V through a resistance lMΩ. The voltage across C after 1 second is approximately**

- 56 V
- 7.8 V
- 6.3 V
- 10 V

**Answer:** 3. 6.3 V

When a capacitor of capacitance C is charged by connecting it in series with a battery of emf E and a resistance R, charge on the capacitor after a time t,

⇒ \(q=C E\left(1-e^{-\frac{t}{R C}}\right)\)

Here, C = lμF, E = 10 V, R = lMΩ

∴ The charge stored on the capacitor after 1 s

\(q=1 \times 10^{-6} \times 10 \times\left(1-e^{\frac{1}{10^6 \times 10^{-6}}}\right)\)= 10^{(-5)(1-e-1)}

= 0.632 x 10^{-5} C

Hence the voltage across the capacitor after 1 s

⇒ \(\frac{q}{C}=\frac{0.632 \times 10^{-5}}{1 \times 10^{-6}}=6.32 \mathrm{~V} \approx 6.3 \mathrm{~V}\)

The option 3 is correct.

**Question 65. Three capacitors of capacitance 1.0μF, 2.0μF, and 5.0μF are connected in series to a 10V source. The potential difference across the 2.0μF capacitors**

- \(\frac{100}{17}\)V
- \(\frac{20}{17}\)V
- \(\frac{50}{17}\)V
- 10V

**Answer:** 3. \(\frac{50}{17}\)V

If the equivalent capacitance of the capacitors is Ceq then,

⇒ \(\frac{1}{C_{\text {eq }}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{5}=\frac{17}{10} \quad\)

or, \(C_{\text {eq }}=\frac{10}{17} \mu \mathrm{F}\)

Charge storedin each capacitor, \(Q=C_{\mathrm{eq}} V=\frac{10}{17} \times 10 \mu \mathrm{C}\)

The potential difference between 2.0μF capacitor

⇒ \(\frac{10 \times 10}{17 \times 2}=\frac{50}{17} \mathrm{~V}\)

The option 3 is correct

**Question 66. An electric bulb, a capacitor, a battery, and a switch are all in series in a circuit How does the intensity of light vary when the switch is turned on?**

- Continues to increase gradually
- Gradually increases for some time and then becomes steady
- Sharply raises initially and then gradually decreases
- Gradually increases for some time and then gradually decreases

**Answer:** 3. Sharply rises initially and then gradually decreases

The given CR circuit with a battery as the voltage source.

If battery voltage = E, resistance ofbulb= R, capacitance= C, then the instantaneous current in the circuit when the switch is turned on,

⇒ \(I(t)=\frac{E}{R} e^{-\frac{t}{R C}}\)

Hence when the switch is turned on, the intensity of the light from the bulb sharply increases initially and then decreases gradually with time.

The option 3 is correct

**Question 67. The insulated plates of a charged parallel plate capacitor (with a small separation between the plates) are approaching each other due to electrostatic attraction. Assuming no other force to be operative and no radiation taking place, which of the following graphs approximately shows the variation with time (t) of the potential difference (V) between the plates?**

**Answer:** 1

When the plates approach each other, the electric field remains constant in the region between the plates, but the voltage between the plates changes.

∴ From E = \(\frac{dV}{dv}\) we get, E x d = V

[d = distance between the plates, V” = potential difference between the plates and £ = electric field in the region between the plates]

E will remain the same, so V ∝ d

The electric force on each plate

⇒ \(F_e=\frac{q \times q}{2 \epsilon_0 A}\)

[Let, A = area of each plate, q = amount of charge]

∴ Acceleration of the plates

⇒ \(a=\frac{q^2}{2 A \epsilon_0 m}\) [m = mass ofeach plate]

So, the distance-time graphic case of uniform acceleration is shown in the figure. As V ∝ d the graph of V vs t is also similar.

The option 1 is correct

**Question 68. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of 2.2 between them. When the electric field in the dielectric is 3 x 10 ^{-4 }V/m, the charge density of the positive plate will be close to**

- 6 x 10
^{4}C/m^{2} - 6 X 10
^{-7}C/m^{2} - 3 x 10
^{-7}C/m^{2} - 3 x 10
^{-4}C/m^{2}

**Answer:** 2. 6 x 10^{-7} C/m^{2}

Electric field, \(E=\frac{\sigma}{\kappa \epsilon_0}\)

∴ Charge density of the plate,

σ =k∈_{0}E = 2.2 x 8.85 x 10^{-12} x 3 x 10^{4}

= 5.48 x 10^{-7} ≈ 6 x 10^{-7}C/m^{2}

The option 2 is correct.

**Question 69. In the given circuit, charge Q _{2} on the 2μF capacitor changes as C is varied from 1μF to 3μF. Q_{2} as a function of C is given properly by (Figures are drawn schematically and are not to scale)**

**Answer:** 2

Equivalent capacitance of the 1μF and 2μF capacitors

= (1+2)

= 3μF

Then, equivalent capacitance ofthe circuit = \(\frac{3 C}{3+C}\)

Charge stored, \(Q=E \cdot \frac{3 C}{3+C}\)

then \(Q_2=\frac{2}{2+1} Q=\frac{2 E C}{C+3}=\frac{2 E}{1+\frac{3}{C}}\)

When C = 1μF, Q_{2} = \(\frac{E}{2}\)

when C = 3μF, Q_{2} = E; the value of Q_{2} increases.

For the mean value (2μF) ofthe capacitors, Q_{2} = \(\frac{4E}{5}\)

Mean value of \(\frac{E}{2}\) and E = \(\frac{\frac{E}{2}+E}{2}=\frac{3 E}{4}<\frac{4 E}{5}\)

Hence, the increase of Q_{2} with C is not linear; there is comparatively greater of Q_{2} in the beginning.

The option 2 is correct

**Question 70. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field due to a point charge Q (having a charge equal to the sum of the charges on the 4μF and 9μF capacitors), at a point of distance 30 m from it, would equal**

- 240 N/C
- 360 N/C
- 420 N/C
- 480 N/C

**Answer:** 3. 420 N/C

The circuit can be simplified

The charge stored on the 3μF capacitor in the second circuit = 3 x 8

= 24μC

So the same charge (24μC) will be stored on the 4μF and 12μF capacitors in the first circuit. Again 12μF is the equivalent capacitance for the 3μF and 9μF capacitors in parallel. If the charge stored on the 9μF capacitor is q, then

⇒ \(\frac{q}{9}=\frac{24-q}{3} \text { or, } q=18 \mu \mathrm{C}\)

∴ Q = 24 + 18

= 42μC

Hence the electric field at a distance of 30 m from Q charge

⇒ \(\frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}, \text { where } r=30 \mathrm{~m}\)

⇒ \(9 \times 10^9 \times \frac{42 \times 10^{-6}}{30^2}=420 \mathrm{~N} / \mathrm{C}\)

The option 3 is correct.

**Question 71. A capacitance of 2.0 pF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of lpF capacitors are available that can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is**

- 2
- 16
- 24
- 32

**Answer:** 4. 32

⇒ \(\frac{1.0 \mathrm{kV}}{300 \mathrm{~V}}=\frac{1000}{300}=3.33\)

∴ The minimum number of capacitor combinations that should be connected in series is 4.

If the capacitance of each combination is x, then

⇒ \(\frac{4}{x}=\frac{1}{2}\) [since equivalent capacitance is 2μF]

∴ x = 8μF

Hence each combination of capacitors must contain 8 1μF capacitors connected in parallel.

∴ Minimum number of capacitors required

=8 x 4

= 32

The option 4. is correct

**Question 72. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant k = \(\frac{5}{3}\) is inserted between the plates, the magnitude of the induced charge will be**

- 2.4 nC
- 0.9 nC
- 1.2 nC
- 0.3 nC

**Answer:** 3. 1.2 nC

⇒ \(Q_i=C V ; Q_f=k C V\)

∴ \(Q_{\text {induced }}=Q_f-Q_i\)

=(k-1)CV

⇒ \(\left(\frac{5}{3}-1\right) \times 90 \times 10^{-12} \times 20\)

= 1.2 x 10^{-9}C

= 1.2 nC

The option 3 is correct.

**Question 73. Two thin dielectric slabs of dielectric constants k _{1} and k_{2} (K_{1} < K_{2}) is inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field F. between the plates with distance d ns measured from plate P Is correctly shown by**

**Answer:** Option 3 is correct

Electric field, \(E \propto \frac{1}{K}\)(k = dielectric constant)

k_{1} < k_{2}

∴ E_{1} > E_{2}

The option 3 is correct

**Question 74. A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?**

- The potential difference between the plates decreases K times
- The energy stored in the capacitor decreases K times
- The change in energy stored is \(\frac{1}{2} C V^2\left(\frac{1}{K}-1\right)\)
- The charge on the capacitor is not conserved

**Answer:** 4. The charge on the capacitor is not conserved

If Q is the charge on the plate,

potential difference \((V)=\frac{Q}{C} ;\)

stored energy \((U)=\frac{1}{2} C V^2=\frac{1}{2} \frac{Q^2}{C}\)

After the dielectric slab is inserted,

capacitance, C’ = KC;

potential difference = \(\frac{Q}{kC}\)

i.e., the potential difference would decrease K times.

Again, stored energy,

⇒ \(U^{\prime}=\frac{1}{2} \frac{Q^2}{C^{\prime}}=\frac{1}{2} \frac{Q^2}{k C}\)

i.e, the stored energy also decreases k times,

Now, change stored energy

⇒ \(=U^{\prime}-U=\frac{1}{2} \frac{Q^2}{C}\left(\frac{1}{K}-1\right)=\frac{1}{2} C V^2\left(\frac{1}{K}-1\right)\)

Hence, options 1, 2, and 3 are correct

But the charge on the capacitor remains conserved in isolating conditions.

The option 4 is correct

**Question 75. A capacitor of 2μF Is charged as shown In the diagram. When the switch S Is turned to position 2, the percentage of Its stored energy dissipated is**

- 20%
- 75%
- 80%
- 0%

**Answer:** 3. 80%

The energy stored in the 2μF capacitor initially,

⇒ \(U_i=\frac{1}{2} \times 2 \times V^2=V^2\)

When the switch is turned to position 2, let V’ be the

voltage across each capacitor.

Total charge is conserved, so \(2 V=2 V^{\prime}+8 V^{\prime} \text { or, } V^{\prime}=\frac{V}{5}\)

Hence, the final energy storedin the two capacitors,

⇒ \(U_f=\frac{1}{2} \times 2 \times \frac{V^2}{25}+\frac{1}{2} \times 8 \times \frac{V^2}{25}=\frac{V^2}{5}\)

∴ Percentage of stored energy dissipated

⇒ \(=\frac{U_i-U_f}{U_f} \times 100 \%=\frac{V^2-\frac{V^2}{5}}{V^2} \times 100 \%=80 \%\)

The option 3 is correct

**Question 76. A parallel plate capacitor is to be designed, using a dielectric of dielectric constant 5, so as to have a dielectric strength of 10 ^{9} V.m^{-1}. If the voltage rating of the capacitor is 12 kV, the minimum area of each plate required to have a capacitance of 80 pF is,q**

- 10.5 x 10
^{-6}m^{2} - 21.7 x 10
^{-6}m^{2} - 25.0 x 10
^{-5}m^{2} - 12.5 x 10
^{-5}m^{2}

**Answer:** 2. 21.7 x 10^{-6} m^{2}

Capacitance, \(C=\frac{\epsilon_0 k A}{d} \quad \text { or, } A=\frac{C d}{\epsilon_0 k}\)

Given, \(k=5 ; E=\frac{V}{d} \quad \text { or, } d=\frac{V}{E}\)

∴ \(A=\frac{C V}{\epsilon_0 k E}=\frac{\left(80 \times 10^{-12}\right) \times\left(12 \times 10^3\right)}{\left(8.85 \times 10^{-12}\right) \times 5 \times 10^9}\)

= 21.7 c 10^{-6}m^{2}

The option 2 is correct

**Question 77. The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is**

- Proportional to the square root of the distance between the plates
- Linearly proportional to the distance between the plates
- Independent of the distance between the plates
- Inversely proportional to the distance between the plates

**Answer:** 3. Independent of the distance between the plates

Let the distance between the two plates of the parallel plate capacitor = d

The electrostatic force between the metal plates

⇒ \(F=Q E=Q \cdot \frac{\sigma}{2 \epsilon_0}=Q \cdot\left(\frac{Q}{A}\right) \cdot \frac{1}{2 \epsilon_0}=\frac{Q^2}{2 A \epsilon_0}\)

Hence the electrostatic force does not depend on the distance between the plates.

The option 3 is correct.

**Question 78. A capacitor C _{1} of capacitance 5μF Is charged to a potential of 100 V and another capacitor C_{2} of capacitance 8μF is charged to 50 V. The positive and negative plates are Mutually connected.**

**1. The final potential of the combination of the two capacitors will be**

- \(\frac{500}{3}\)V
- \(\frac{900}{3}\)V
- 150V
- 50V

**Answer:** 2. \(\frac{900}{3}\)V

**2. The amount of charge of the capacitor C _{1} after combination will be**

- \(\frac{4500}{13} \mu \mathrm{C}\)
- \(\frac{7200}{13} \mu \mathrm{C}\)
- \(\frac{2700}{13} \mu \mathrm{C}\)
- \(\frac{11700}{13} \mu \mathrm{C}\)

**Answer:** 1. \(\frac{4500}{13} \mu \mathrm{C}\)

**3. The amount of charge of the capacitor C _{2} after combination will be**

- 4500μC
- 7200μC
- \(\frac{4500}{13} \mu \mathrm{C}\)
- \(\frac{7200}{13} \mu \mathrm{C}\)

**Answer:** 4. \(\frac{7200}{13} \mu \mathrm{C}\)

**4. Energy loss will be**

- 3.11 x 10
^{-13}J - 35 x 10
^{-2}J - 3.9 x 10
^{-13}J - 7.8 x 10
^{-5}J

**Answer:** 3. 3.9 x 10^{-13} J

**Question 79. A parallel plate capacitor of plate area 0.2 m ^{2} and spacing 10^{-2} m is charged to 10^{3} V and is then disconnected from the battery**

**1. If the plates are pulled apart to double the plate spacing capacitance of the capacitor will be**

- 44.25 pF
- 88.5 pF
- 120.45 pF
- 22.12 pF

**Answer:** 2. 88.5 pF

**2. The amount of work required to double the plate spacing is**

- 8.85 x 10
^{-5}J - 17.7 x 10
^{-51}J - 4.42 x 10
^{-5}J - 26.55 x 10
^{-7}J

**Answer:** 1. 8.85 x 10^{-5} J

**3. The final voltage of the capacitor will be**

- 10
^{3}V - 4 x 10
^{3}V - 2 x 10
^{3}V - 10
^{6}V

**Answer:** 3. 2 x 10^{3} V

**Question 80. A spherical drop of water carries a charge of 10 x 10 ^{-12} C and has a potential of 100V at its surface**

**1. The radius of the drop will be**

- 9 x 10
^{-3}m - 9 x 10
^{-5}m - 9 x 10
^{-2}m - 9 x 10
^{-4}m

**Answer:** 4. 9 x 10^{-4} m

**2. If eight such charged drops as mentioned above, combine to form a single drop, the potential at the surface of the new drop will be**

- \(\frac{4}{3}\)
- 400 V
- 200 V
- 300 V

**Answer:** 2. 400 V

**Question 81. Two capacitors of capacity 6μF and 3μF are charged to 100 V and 50 V separately and connected to a 200 V battery using three switches. Now all three switches S _{1}, S_{2,} and S_{3} are closed.**

**1. Which plates form an isolated system?**

- Plates 1 and 4 separately
- Plates 2 and 3 separately
- Plates 1 and 4 jointly
- Plates 2 and 3 jointly

**Answer:** 4. Plates 2 and 3 jointly

**2. Charges on 6μF and 3μF capacitors in steady state** **will be**

- 400μC, 400μC
- 700μC, 250μC
- 800μC, 350μC
- 300μC, 450μC

**Answer:** 2. 700μC, 250μC

**3. Suppose q _{1}, q_{2,} and q_{3} are the magnitudes of charges that flow through switches S_{1}, S_{2} and S_{3} after they are closed. Then**

- q
_{1}= q_{3}and q_{2}= 0 - q
_{1}= q_{3}= \(\frac{q_2}{2}\) - q
_{1}= q_{3}= 2q_{2} - q
_{1}= q_{2}= q_{3}

**Answer:** 4. q_{1} = q_{2} = q_{3}