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

Q 11 :

A parallel plate capacitor has plate area $40 {cm}^{2}$ and plate separation 2 mm. The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 5. The capacitance of the system is [2023]

$\frac{10}{3} ε_{0} F$
$\frac{3}{10} ε_{0} F$
$24 ε_{0} F$
$10 ε_{0} F$

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$A = 40 \times 10^{- 4} m^{2} = 4 \times 10^{- 3} m^{2}$

$d = 2 \times 10^{- 3} m, k = 5, t = 10^{- 3} m, C = ?$

$C = \frac{ε_{0} A}{d - t + \frac{t}{k}}$ $= \frac{ε_{0} \times 4 \times 10^{- 3}}{(2 - 1) \times 10^{- 3} + \frac{10^{- 3}}{5}}$ $= \frac{4 ε_{0} \times 5}{6} = \frac{10 ε_{0}}{3} F$

Q 12 :

The distance between two plates of a capacitor is $d$ and its capacitance is $C_{1}$ , when air is the medium between the plates. If a metal sheet of thickness $\frac{2 d}{3}$ and of the same area as the plates is introduced between the plates, the capacitance of the capacitor becomes $C_{2}$ . The ratio $\frac{C_{2}}{C_{1}}$ is [2023]

2 : 1
4 : 1
3 : 1
1 : 1

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$K_{metal sheet} = \infty, t = \frac{2 d}{3}$

$C_{1} = \frac{ε_{0} A}{d}$

$C_{2} = \frac{ε_{0} A}{d - t + \frac{t}{k}} = \frac{ε_{0} A}{d - \frac{2 d}{3} + 0} = 3 C_{1}$

$\frac{C_{2}}{C_{1}} = 3$

Q 13 :

A parallel plate capacitor with air between the plates has a capacitance of 15 pF. The separation between the plates becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes $\frac{x}{4} pF$ . The value of $x$ is _________. [2023]

0%

(105)

$C_{0} = \frac{ε_{0} A}{d} = 15 pF$

$C = \frac{K ε_{0} A}{2 d} = \frac{3.5}{2} \times 15 pF = \frac{105}{4} pF$

Q 14 :

A capacitor has capacitance $5 μ F$ when its parallel plates are separated by air medium of thickness $d$ . A slab of material of dielectric constant 1.5 having area equal to that of the plates but thickness $\frac{d}{2}$ is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be _________ $μ F$ . [2023]

0%

(6)

$\frac{ε_{0} A}{d} = 5 μ F$

$C_{new} = \frac{ε_{0} A}{\frac{(\frac{d}{2})}{1.5} + \frac{(\frac{d}{2})}{1}}$

$= \frac{ε_{0} A}{(\frac{d}{3} + \frac{d}{2})} = \frac{6 ε_{0} A}{5 d} = \frac{6}{5} \times 5 μ F = 6 μ F$

Q 15 :

Two parallel plate capacitors $C_{1}$ and $C_{2}$ , each having capacitance of $10 μ F$ , are individually charged by a 100 V D.C. source. Capacitor $C_{1}$ is kept connected to the source and a dielectric slab is inserted between its plates. Capacitor $C_{2}$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards, the capacitor $C_{1}$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be ____ V. (Assuming dielectric constant =10) [2023]

0%

(55)

$Charge on C_{1} = K C E$

$and charge on C_{2} = C E$

When they are connected in parallel, charge will be equally divided, so charge on one capacitor is

$q = \frac{K + 1}{2} C V$

$So$ $V = \frac{q}{K C} = \frac{K + 1}{2 K} = 55 V$

Q 16 :

A parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant K = 4. The thickness of the dielectric material is $x$ . where $x < d$ .

Let $C_{1}$ and $C_{2}$ be the capacitance of the system for $x = \frac{1}{3} d$ and $x = \frac{2 d}{3}$ respectively. If $C_{1} = 2 μ F$ , the value of $C_{2}$ is ________ $μ F$ . [2023]

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(3)

$For x = \frac{d}{3}$

$C_{1} = \frac{ε_{0} A}{(\frac{d / 3}{k} + \frac{2 d}{3})} = \frac{ε_{0} A}{\frac{d}{12} + \frac{2 d}{3}} = \frac{ε_{0} A}{d} \times \frac{12}{9}$

$C_{1} = \frac{4}{3} \frac{ε_{0} A}{d} = 2 μ F$

$For x = \frac{2 d}{3}$

$C_{2} = \frac{ε_{0} A}{(\frac{2 d / 3}{k} + \frac{d}{3})} = \frac{ε_{0} A}{d} \times 2 = \frac{6}{4} \times 2 = 3 μ F$

Q 17 :

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness ${(1 / 3)}^{rd}$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is : [2026]

$\frac{C K}{2 + K}$
$\frac{3 C K^{2}}{{(2 K + 1)}^{2}}$
$\frac{3 K C}{2 K + 1}$
$\frac{4 K C}{3 K - 1}$

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[IMAGE 87]----------

$C_{1} = \frac{3 A ε_{0}}{2 d}$ $C_{2} = \frac{3 A ε_{0} \times K}{d}$

$C_{1} = \frac{3}{2} C$ $C_{2} = 3 K C$

$C_{eq} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{\frac{3}{2} C \times 3 K C}{\frac{3}{2} C + 3 K C}$

$C_{eq} = \frac{\frac{9}{2} K C^{2}}{\frac{3}{2} C (2 K + 1)} = \frac{3 K C}{2 K + 1}$

Q 18 :

Identify the correct statements:

A. Effective capacitance of a series combination of capacitors is always smaller than the smallest capacitance of the capacitor in the combination.

B. When a dielectric medium is placed between the charged plates of a capacitor, displacement of charges cannot occur due to insulation property of dielectric.

C. Increasing of area of capacitor plate or decreasing of thickness of dielectric is an alternate method to increase the capacitance.

D. For a point charge, concentric spherical shells centered at the location of the charge are equipotential surfaces.

Choose the correct answer from the options given below: [2026]

A, C and D Only
B and D Only
C and D Only
A, B and C Only

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Q 19 :

A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws 25% more charge from the battery. The dielectric constant of mica is __________. [2026]

2.0
1.5
1.0
2.5

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Topic Question Set

Q 11 :

A parallel plate capacitor has plate area $40 {cm}^{2}$ and plate separation 2 mm. The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 5. The capacitance of the system is [2023]

0%

0%

0%

0%

Q 17 :

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness ${(1 / 3)}^{rd}$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is : [2026]

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