Electric Potential and Capacitance | Physics JEE previous year questions with Solutions

Q 1 :
A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab $(ε_{r} = 6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _______ $\times 10^{- 12} J .$ [2024]

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(750) Before inserting dielectric, capacitance is given as $C_{0} = 12.5$ pF and charge on the capacitor is $Q = C_{0} V$ . After inserting dielectric capacitance will become $C_{f} = ε_{r} C_{0} .$

Change in potential energy of the capacitor

$∆ U = U_{0} - U_{f} = \frac{Q^{2}}{2 C_{0}} - \frac{Q^{2}}{2 C_{f}} = \frac{Q^{2}}{2 C_{0}} [1 - \frac{1}{ε_{r}}]$

$∆ U = \frac{{(C_{0} V)}^{2}}{2 C_{0}} [1 - \frac{1}{ε_{r}}] = \frac{1}{2} C_{0} V^{2} [1 - \frac{1}{ε_{r}}]$

Using $C_{0} = 12.5 p F,$ $V = 12$ $V,$ $ε_{r} = 6$

$∆ U = \frac{1}{2} (12.5) \times 12^{2} [1 - \frac{1}{6}] = \frac{1}{2} (12.5) \times 12^{2} \times \frac{5}{6}$

$= 750 p J = 750 \times 10^{- 12} J$

Q 2 :
A parallel plate capacitor with plate separation 5mm is charged up by a battery. It is found that on introducing a dielectric sheet of thickness 2 mm, while keeping the battery connections intact, the capacitor draws 25% more charge from the battery than before. The dielectric constant of the sheet is __________. [2024]

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(2) Without dielectric $Q = \frac{A \in_{0}}{d} V$

with dielectric

$Q = \frac{A \in_{0} V}{d - t + \frac{t}{K}}$

Given $\frac{A \in_{0} V}{d - t + \frac{t}{K}} = (1.25) \frac{A \in_{0} V}{d}$

$\Rightarrow 1.25 (3 + \frac{2}{K}) = 5 \Rightarrow K = 2$

Q 3 :
A capacitor has air as dielectric medium and two conducting plates of area 12 ${cm}^{2}$ and they are 0.6 cm apart. When a slab of dielectric having area 12 ${cm}^{2}$ and 0.6 cm thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm to keep the capacitance same as in previous case. The dielectric constant of the slab is

(Given $ε_{0} = 8.834 \times 10^{- 12}$ F/m) [2024]

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0.66
1.33
1.50

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(4)

Parallel plate capacitance, $C = \frac{ϵ_{o} A}{d_{1}}$

For partially filled capacity, $C = \frac{ϵ_{o} A}{(d_{2} - t + \frac{t}{k})}$

For case (i) $C_{1} = \frac{ϵ_{o} A}{0.6}$ , For case (ii) $C_{2} = \frac{ϵ_{o} A}{0.8 - 0.6 + \frac{0.6}{K}}$

Now $C_{1} = C_{2}, \frac{ϵ_{o} A}{0.6} = \frac{ϵ_{o} A}{0.2 + \frac{0.6}{K}}$ ,

$0.6 = 0.2 + \frac{0.6}{K} \Rightarrow K = \frac{3}{2}$

Q 4 :
Consider a parallel plate capacitor of area A (of each plate) and separation $' d'$ between the plates. If E is the electric field and $ε_{0}$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is [2025]

$\frac{1}{2} ε_{0} E^{2} A d$
$\frac{3}{4} ε_{0} E^{2} A d$
$\frac{1}{4} ε_{0} E^{2} A d$
$ε_{0} E^{2} A d$

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(1)

We know energy density

$ρ_{av} = \frac{1}{2} ε_{0} E^{2}$

So potential energy $= ρ_{av} \times$ Volume

$\Rightarrow P . E . = \frac{1}{2} ε_{0} E^{2} A d$

Q 5 :
A parallel plate capacitor was made with two rectangular plates, each with a length of $l$ = 3 cm and breath of b = 1 cm. The distance between the plates is 3 $μ$ m. Out of the following, which are the ways to increase the capacitance by a factor of 10?

A. $l$ = 30 cm, b = 1 cm, d = 1 $μ$ m

B. $l$ = 3 cm, b = 1 cm, d = 30 $μ$ m

C. $l$ = 6 cm, b = 5 cm, d = 3 $μ$ m

D. $l$ = 1 cm, b = 1 cm, d = 10 $μ$ m

E. $l$ = 5 cm, b = 2 cm, d = 1 $μ$ m

Choose the correct answer from the options given below: [2025]

C and E only
B and D only
A only
C only

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(1)

We know $C = \frac{A ε_{0}}{d} = \frac{b l ε_{0}}{d}$

So, to increase the capacitance by factor $10 (\frac{A}{d})$ has to increase by factor 10.

For option (A) $C' = 30 C$

For option (B) $C' = \frac{C}{10}$

For option (C) $C' = 10 C$

For option (D) $C' = \frac{C}{10}$

For option (E) $C' = 10 C$

Q 6 :
A parallel plate capacitor of capacitance 1 $μ$ F is charged to a potential difference of 20 V. The distance between plates is 1 $μ$ F. The energy density between plates of capacitor is : [2025]

$1.8 \times 10^{3} J / m^{3}$
$2 \times 10^{4} J / m^{3}$
$2 \times 10^{2} J / m^{3}$
$1.8 \times 10^{5} J / m^{3}$

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(1)

Energy density $u = \frac{1}{2} ε_{0} E^{2} = \frac{1}{2} ε_{0} {(\frac{V}{d})}^{2}$

$\Rightarrow u = \frac{1}{2} (8.85 \times 10^{- 12}) {(\frac{20}{10^{- 6}})}^{2} J / m^{3} = 1.8 \times 10^{3} J / m^{3}$

Q 7 :
A parallel-plate capacitor of capacitance 40 $μ$ F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are [2025]

2 mC and 0.2 J
8 mC and 2.0 J
4 mC and 0.2 J
2 mC and 0.4 J

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(3)

$C = 40 μ F, C' = k C = 80 μ F$

V = 100 V

Q = CV = 4 mC

$Q' = C' V = 8 mC$

$∴$ Extra charge, $△ Q$ = 4 mC

$U_{1} = \frac{1}{2} {CV}^{2}, U_{2} = \frac{1}{2} C' V^{2}$

$△ U = \frac{1}{2} (C' - C) V^{2} = \frac{1}{2} \times 40 \times 10^{4} \times 10^{- 6} = 0.2 J$

Q 8 :
A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $ε_{1}$ and $ε_{2}$ , as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are $C_{1}$ and $C_{2}$ respectively, then $\frac{C_{1}}{C_{2}}$ is [2025]

$\frac{ε_{1} ε_{2}^{2}}{{(ε_{1} + ε_{2})}^{2}}$
$\frac{4 ε_{1} ε_{2}}{{(ε_{1} + ε_{2})}^{2}}$
$\frac{ε_{1} ε_{2}}{ε_{1} + ε_{2}}$
$\frac{ε_{0} (ε_{1} + ε_{2})}{2}$

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(2)

$C_{1} = \frac{ε_{0} A}{\frac{d}{2 ε_{1}} + \frac{d}{2 ε_{2}}} = \frac{ε_{0} A}{d} {\frac{1}{\frac{ε_{2} + ε_{1}}{2 ε_{1} ε_{2}}}}$

$C_{2} = \frac{ε_{0}}{d} (ε_{1} \frac{A}{2} + ε_{2} \frac{A}{2}) = \frac{ε_{0} A}{d} {\frac{ε_{1}}{2} + \frac{ε_{2}}{2}}$

$\frac{C_{1}}{C_{2}} = \frac{2 ε_{1} ε_{2}}{(ε_{2} + ε_{1}) \frac{(ε_{1} + ε_{2})}{2}} = \frac{4 ε_{1} ε_{2}}{{(ε_{1} + ε_{2})}^{2}}$

Q 9 :
Three parallel plate capacitors $C_{1}, C_{2}$ and $C_{3}$ each of capacitance 5 $μ$ F are connected as shown in figure. The effective capacitance between points A and B, when the space between the parallel plates of $C_{1}$ capacitor is filled with a dielectric medium having dielectric constant of 4, is [2025]

22.5 $μ F$
7.5 $μ F$
9 $μ F$
30 $μ F$

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(3)

$C_{eq} = (\frac{K C \cdot C}{K C + C}) + C = (\frac{K}{K + 1}) C + C$

$\Rightarrow C_{eq} = \frac{4}{5} \times 5 + 5 = 9 μ F$

Q 10 :
A parallel plate capacitor has charge $5 \times 10^{- 6} C$ . A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{- 6} C$ then the dielectric constant of the slab is __________ [2025]

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(5)

$E_{net} = E_{0} - E_{i n}$

$E_{in} = E_{0} (1 - \frac{1}{k})$

$Q_{in} = Q_{0} (1 - \frac{1}{k})$

$4 \times 10^{- 6} = 5 \times 10^{- 6} (1 - \frac{1}{k}) \Rightarrow k = 5$

Topic Question Set

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Q 4 :
Consider a parallel plate capacitor of area A (of each plate) and separation $' d'$ between the plates. If E is the electric field and $ε_{0}$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is [2025]

Q 6 :
A parallel plate capacitor of capacitance 1 $μ$ F is charged to a potential difference of 20 V. The distance between plates is 1 $μ$ F. The energy density between plates of capacitor is : [2025]

Q 10 :
A parallel plate capacitor has charge $5 \times 10^{- 6} C$ . A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{- 6} C$ then the dielectric constant of the slab is __________ [2025]

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Q 4 : Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If E is the electric field and ε0 is the permittivity of free space between the plates, then potential energy stored in the capacitor is [2025]

Q 6 : A parallel plate capacitor of capacitance 1 μF is charged to a potential difference of 20 V. The distance between plates is 1 μF. The energy density between plates of capacitor is : [2025]

Q 9 : Three parallel plate capacitors C1,C2 and C3 each of capacitance 5 μF are connected as shown in figure. The effective capacitance between points A and B, when the space between the parallel plates of C1 capacitor is filled with a dielectric medium having dielectric constant of 4, is [2025]

Q 10 : A parallel plate capacitor has charge 5×10–6C. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4×10–6C then the dielectric constant of the slab is __________ [2025]

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Q 4 :
Consider a parallel plate capacitor of area A (of each plate) and separation $' d'$ between the plates. If E is the electric field and $ε_{0}$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is [2025]

Q 6 :
A parallel plate capacitor of capacitance 1 $μ$ F is charged to a potential difference of 20 V. The distance between plates is 1 $μ$ F. The energy density between plates of capacitor is : [2025]

Q 10 :
A parallel plate capacitor has charge $5 \times 10^{- 6} C$ . A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{- 6} C$ then the dielectric constant of the slab is __________ [2025]