Question

When a charge of $10\mu \text{C}$ is shifted from infinity to a point in an electric field, it is found that work done by electrostatic force is $-10\mu \text{J}$. If the charge is doubled and taken again from infinity to the same point without accelerating it, then work done by electrostatic force will be

• A. $-10\mu \text{J}$
• B. $-20\mu \text{J}$
• C. $+10\mu \text{J}$
• D. $+20\mu \text{J}$

Answer 1

The correct option is B $-20\mu \text{J}$

We know the relation between work done by the external force and the potential at any point $A\text{(say)}$ in an electric field,
$V_A=\frac{W_{\text{ext}(\infty \to A)}}{q_o}$
Also, $W_{\text{ext}(\infty \to A)}=-W_{\text{elect}(\infty \to A)}$
So, $V_A=\frac{-W_{\text{elect}(\infty \to A)}}{q_o}$
As the point in the electric field is the same, therefore, the potential will also be the same.
So, $W_{\text{elect}(\infty \to A)}\propto q_o$

If the charge is doubled, the work done by the electrostatic force will also be doubled.
Hence, work done by electrostatic force will be $-20\mu \text{J}$.