Two identical thin metal plates has charge $\mathrm{q}_{1}$ and $\mathrm{q}_{2}$ respectively such that $\mathrm{q}_{1}>\mathrm{q}_{2}$. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

(C) $\frac{\left(q_{1}-q_{2}\right)}{2 C}$

(A) $\frac{\left(q_{1}+q_{2}\right)}{C}$

(B) $\frac{\left(q_{1}-q_{2}\right)}{C}$

(D) $\frac{2\left(q_{1}-q_{2}\right)}{C}$

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Capacitance

Two identical thin metal plates has charge

q_{1}

and

q_{2}

respectively such that

q_{1} > q_{2}

. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

(A) $\frac{( q _{1} + q _{2} )}{C}$
(B) $\frac{( q _{1} - q _{2} )}{C}$
(C) $\frac{( q _{1} - q _{2} )}{2 C}$ ✓ Correct
(D) $\frac{2 ( q _{1} - q _{2} )}{C}$

Answer: (C)

Solution

When two charged plates are brought close together, the electric field between them is due to the net charge difference.

The electric field between the plates:

E = \frac{q _{1} - q _{2}}{2 A ϵ _{0}}

The potential difference:

V = E d = \frac{q _{1} - q _{2}}{2 A ϵ _{0}} \cdot d

Since capacitance

C = \frac{A ϵ _{0}}{d}

, we have

A ϵ_{0} = C d

Substituting:

V = \frac{q _{1} - q _{2}}{2 C d} \cdot d = \frac{q _{1} - q _{2}}{2 C}

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Capacitance

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