Question

Two conducting plates of surface area $5{\text{m}}^2$ are placed parallel to each other at a distance $0.885\text{cm}$. If the charge on one plate is $60\mu \text{C}$ and on the other is $20\mu \text{C}$, find the potential difference between the plates.

• A. $4\text{V}$
• B. $4\text{kV}$
• C. $40\text{V}$
• D. $40\text{kV}$

Answer 1

The correct option is B $4\text{kV}$

Charge given to the plates are $60\mu \text{C}$ and $20\mu \text{C}$

This means the charge on outer side of the plates will be

$q=\frac{60+20}{2}=40\mu \text{C}$


Hence, charge on the inner plates of the capacitor is $60-40=20\mu \text{C}$ and $20-40=-20\mu \text{C}$

Now, we know that $Q=CV$

$\Rightarrow Q=\frac{{ϵ}_{0}A}{d}V$

$\Rightarrow 20\times {10}^{-6}=\frac{8.85\times {10}^{-12}\times 5}{0.885\times {10}^{-2}}V$

$\Rightarrow V=4\text{kV}$

Hence, option $\left(b\right)$ is correct.