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 K1 is closed and K2 is open. Then K1 is opened and K2 is closed(order is important). Take Q1 and Q2 as charges on C1 and C2 and V1 and V2 as voltage respectively. Then
Read and Learn More Class 12 Physics Multiple Choice Questions
- Charge on C1 gets redistributed such that V1 = V2
- Charge on C1 gets redistributed such that Q’1 = Q2
- Charge on C1 gets redistributed such that C1V1 + C2V2 = C1E
- Charge on C1 gets redistributed such that Q’2+Q’2 = Q
Answer:
1. Charge on C1 gets redistributed such that V1 = V2
4. Charge on C1 gets redistributed such that
When K1 is closed and K2 is open, Q (on C1) =EC1.
When K2 is closed and K1 is open, there being no battery in the circuit, C1 and C2 are connected in parallel and the charge on C1 will be redistributed between C1 and C2 in such a manner that V1 = V2.
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
- n1/3V
- n2/3V
- n3/2V
- n3V
Answer: 2. n2/3V
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 r1 and r2 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 C1 and C2 (= 2C1) are connected in a circle with a switch. Initially, the switch is open and C1 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
- k2
- 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 C1 and C2 are connected to a battery. C1 is filled with air and C2 is filled with an insulator of dielectric constant ∈r then
- Q1 > Q2
- Q1 < Q2
- Q1 = Q2
- None
Answer: 2. Q1 < Q2
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 C1 and C2, 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 C1 when four capacitors of equal capacitance are connected in series. In their parallel connection, the equivalent capacitance becomes C2. The ratio C1/C2 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)CV2
- 0
Answer: 4. 0
Question 40. The work done to charge a capacitor of capacitance 100μF with 8 x 10-18C will be
- 32 X 10-32J
- 16 X 10-32J
- 3.2 X 10-26J
- 4 X 10-10J
Answer: 1. 32 X 10-32J
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
- k2
- 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
- x2
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
- CAB = 4μF
- \(C_{A B}=\frac{18}{11} \mu \mathrm{F}\)
- V = 20 V
- V = 40 V
Answer:
1. CAB = 4μF
4. V = 40 V
Question 53. In the given circuit C1 = C2 = 2μF. Then charge is stored in
- Capacitor C1 is zero
- Capacitor C2 is zero
- Both C1 and C2 is zero
- Capacitor C1 is 40μC
Answer:
2. Capacitor C2 is zero
4. Capacitor C1 is 40μC
Question 54. In the given network which one is correct?
- \(\left|q_2\right|=280 \mu \mathrm{C}\)
- \(\left|q_3\right|=160 \mu \mathrm{C}\)
- q2 = 120μC, q3 = 0
- Impossible to find q2 and q3 unless C1 and Q2 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 C1 is charged up to potential V and then connected in parallel to an uncharged capacitor capacitance C2. 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 = C1 V When the charged capacitor C1 is connected in parallel to the uncharged capacitor C2, then the equivalent capacitance of the combination, Ceq = C1 + C2. 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 r1 and r2 (r2 > r1). 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, σ1 = k σ2
∴ σ1 ≠ σ2
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
Ceq = 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 104 C/m2
- 6 X 10-7 C/m2
- 3 x 10-7 C/m2
- 3 x 10-4 C/m2
Answer: 2. 6 x 10-7 C/m2
Electric field, \(E=\frac{\sigma}{\kappa \epsilon_0}\)
∴ Charge density of the plate,
σ =k∈0E = 2.2 x 8.85 x 10-12 x 3 x 104
= 5.48 x 10-7 ≈ 6 x 10-7C/m2
The option 2 is correct.
Question 69. In the given circuit, charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF. Q2 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, Q2 = \(\frac{E}{2}\)
when C = 3μF, Q2 = E; the value of Q2 increases.
For the mean value (2μF) ofthe capacitors, Q2 = \(\frac{4E}{5}\)
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 Q2 with C is not linear; there is comparatively greater of Q2 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 10-9C
= 1.2 nC
The option 3 is correct.
Question 73. Two thin dielectric slabs of dielectric constants k1 and k2 (K1 < K2) 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)
k1 < k2
∴ E1 > E2
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 109 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 m2
- 21.7 x 10-6 m2
- 25.0 x 10-5 m2
- 12.5 x 10-5 m2
Answer: 2. 21.7 x 10-6 m2
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 10-6m2
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 C1 of capacitance 5μF Is charged to a potential of 100 V and another capacitor C2 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 C1 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 C2 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 m2 and spacing 10-2 m is charged to 103 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
- 103V
- 4 x 103 V
- 2 x 103 V
- 106V
Answer: 3. 2 x 103 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 S1, S2, and S3 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 q1, q2, and q3 are the magnitudes of charges that flow through switches S1, S2 and S3 after they are closed. Then
- q1 = q3 and q2 = 0
- q1 = q3 = \(\frac{q_2}{2}\)
- q1 = q3 = 2q2
- q1 = q2 = q3
Answer: 4. q1 = q2 = q3