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

Q 1 :

A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2 C and potential 2 V. Then the loss of energy is $\frac{x}{3} E$ , where $x$ is ______ . [2024]

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(2) Initial potential energy of system

$\Rightarrow \frac{1}{2} C V^{2} + \frac{1}{2} (2 C) {(2 V)}^{2}$

$U_{i} = E + 8 E = 9 E$

Common potential difference across capacitors, finally

$\Rightarrow V_{0} = 5 V / 3$

Final potential energy of system

$U_{f} = \frac{1}{2} C V_{0}^{2} + \frac{1}{2} (2 C) V_{0}^{2}$

$U_{f} = \frac{3 C}{2} V_{0}^{2} = \frac{3 C}{2} {(\frac{5 V}{3})}^{2} = \frac{1}{2} C V^{2} (\frac{25}{3}) = \frac{25}{3} E$

Loss of energy

$\Rightarrow U_{i} - U_{f} = 9 E - \frac{25 E}{3} = \frac{(27 - 25) E}{3} = \frac{2 E}{3}$

Q 2 :

A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A. If the rate of change of potential difference between the plates is $7 \times 10^{8}$ V/s then the integer value of the distance between the parallel plates is $(Take ε_{0} = 9 \times 10^{- 12} \frac{F}{m}, π = \frac{22}{7})______μ m$ [2025]

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(1320)

$V = \frac{Q}{C} = \frac{i t}{(\frac{ε_{0} A}{d})} = \frac{i t d}{ε_{0} (π r^{2})}$

$\Rightarrow d = \frac{ε_{0} (π r^{2})}{i} (\frac{V}{t})$

$d = \frac{(9 \times 10^{- 12}) (\frac{22}{7}) {(0.1)}^{2}}{0.15} (7 \times 10^{8}) m = 1320 μm$

Q 3 :

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one.

Reason R: Capacitance of metallic spheres depend on the radii of spheres.

In the light of the above statements, choose the correct answer from the options given below. [2023]

A is false but R is true
Both A and R are true and R is the correct explanation of A
A is true but R is false
Both A and R are true but R is not the correct explanation of A

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(1)

$Potential of a conducting sphere is$

$V = \frac{K Q}{R} (Solid as well as hollow)$

$V_{1} = V_{2} and R_{1} = R_{2}$

$∴$ $Q_{1} = Q_{2}$

Q 4 :

A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is [2023]

$\frac{C V}{2 ε_{0}}$
$\frac{2 C V}{ε_{0}}$
$\frac{C V}{ε_{0}}$
$zero$

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(3)

$ϕ = \frac{q_{in}}{ε_{0}} = \frac{Q}{ε_{0}} = \frac{C V}{ε_{0}}$

Q 5 :

As shown in the figure, two parallel plate capacitors having equal plate area of $200 {cm}^{2}$ are joined in such a way that $a \neq b$ . The equivalent capacitance of the combination is $x ε_{0} F$ . The value of $x$ is ___________ . [2023]

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(5)

$C = \frac{ε_{0} A}{(d - c)} = \frac{ε_{0} \times 200 \times 10^{- 4}}{4 \times 10^{- 3}}$

$∴$ $x = 5$

$The situation is equivalent to a conducting slab placed between the plates.$

Topic Question Set

Q 1 :

A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2 C and potential 2 V. Then the loss of energy is $\frac{x}{3} E$ , where $x$ is ______ . [2024]

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Q 4 :

A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is [2023]

Q 5 :

As shown in the figure, two parallel plate capacitors having equal plate area of $200 {cm}^{2}$ are joined in such a way that $a \neq b$ . The equivalent capacitance of the combination is $x ε_{0} F$ . The value of $x$ is ___________ . [2023]

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