Question

Two conductors of irregular shapes placed near each other are connected to the two terminals of a battery of $50\text{V}$. It is observed that the charge on one of the conductors is $2\mu \text{C}$. The capacitance of this arrangement is

• A. $2\times {10}^{-8}\text{F}$
• B. $4\times {10}^{-8}\text{F}$
• C. ${10}^{-8}\text{F}$
• D. $4\times {10}^{-6}\text{F}$

Answer 1

The correct option is B $4\times {10}^{-8}\text{F}$

Given:
$Q=2\mu \text{C}=2\times {10}^{-6}\text{C};V=50\text{V}$

An arrangement of two conductors of any shape and size kept close to each other is a capacitor. How effective the capacitor will be is decided by the capacitance of the combination.

Apply formula $Q=CV$, we get

$\Rightarrow C=\frac{Q}{V}$

$\Rightarrow C=\frac{2\times {10}^{-6}\text{C}}{50\text{V}}=4\times {10}^{-8}\text{F}$

So, option $\left(b\right)$ is right.

Key concept:- $Q=CV$ is an equation valid for any two conductors placed near to each other.