Question

A charge $2\mu \text{C}$ is taken from infinity to any point $P\text{(say)}$ in an electric field, without changing its velocity. If workdone against electrostatic forces is $-40\mu \text{J}$, then the electric potential at that point will be:

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

Answer 1

The correct option is D $-20\text{V}$

Let the charge $q\left(2\mu \text{C}\right)$ is taken from $\infty$ to point $\text{P}$, thus the relation for work done by external force is given by

$\frac{W_{\text{ext}(\infty \to P)}}{q}=V_P-V_{\infty }$

Since, we take $V_{\infty }=0$

$\Rightarrow \frac{W_{\text{ext}(\infty \to P)}}{q}=V_P....\left(1\right)$

Since the work done against electrostatic forces is $-40\mu \text{J}$, it suggests that $W_{\text{ext}}=-40\mu \text{J}$

From Eq.$\left(1\right)$,

$\frac{-40\mu \text{J}}{2\mu C}=V_P$

$\therefore V_P=-20\text{V}$

So, $\left(d\right)$ is the correct choice.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: Here work done against electrostatic forces is -ve}\\ \text{thus work done by external force will be -ve.}\\ \text{It indicates that the charge is moved along the direction}\\ \text{of electric field by electric forces. Thus its potential has decreased}(V_i=0,V_f=-20\text{V})\end{array}$