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Capacitance of Parallel Plate Capacitor

Question

Two identical metal plates are given positive charge $Q_{1}$ and $Q_{2}$ $(< Q_{1})$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$ , the potential difference between them is
$(Q_{1} + Q_{2}) / 2 C$
$(Q_{1} + Q_{2}) / C$
$(Q_{1} - Q_{2}) / C$
$(Q_{1} - Q_{2}) / (2 C)$

A

(Q_{1} - Q_{2}) / C

B

(Q_{1} + Q_{2}) / C

C

(Q_{1} + Q_{2}) / 2 C

D

(Q_{1} - Q_{2}) / (2 C)

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Solution

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The potential difference between the two identical metal plates is given as

$C = \frac{ε_{0} A}{d}$

Let the surface charge density is given as

$σ_{1} = σ_{21} = \frac{Q}{A}$

The net electric field is

$E_{n e t} = \frac{σ_{1} - σ_{2}}{2 ε_{0}}$

We know the potential difference is given as

$V = E . d$

By substituting the above values we get

$V = \frac{Q_{1} - Q_{2}}{2 C}$

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