Question

A unit positive charge has to be brought from infinity to a mid-point between two charges, $20\mu \text{C}$ and $10\mu \text{C}$ separated by a distance of $50\text{m}$. How much work will be required by an external force?

• A. $0.1\times {10}^4\text{J}$
• B. $1.1\times {10}^4\text{J}$
• C. $2.1\times {10}^4\text{J}$
• D. $3.1\times {10}^4\text{J}$

Answer 1

The correct option is B $1.1\times {10}^4\text{J}$

Given:
Point charge, $q_o=1\text{C}$

Let point $M$ be the mid-point between the two charges.


At point $M$, total potential, $V_M=V_{q_1}+V_{q_2}$
$\Rightarrow V_M=\frac{k{q}_1}{r_1}+\frac{k{q}_2}{r_2}$
$\Rightarrow V_M=\frac{9\times {10}^9\times 20\times {10}^{-6}}{25}+\frac{9\times {10}^9\times 10\times {10}^{-6}}{25}$
$\Rightarrow V_M=1.1\times {10}^4\text{V}$

We know, $V_M=\frac{W_{\text{ext}(\infty \to M)}}{q_o}$
So, $1.1\times {10}^4=\frac{W_{\text{ext}(\infty \to M)}}{1}$
$\Rightarrow W_{\text{ext}(\infty \to M)}=1.1\times {10}^4\text{J}$

External force will have to do $1.1\times {10}^4\text{J}$ of work.

$\begin{array}{c}\text{Why this question?}\\ \text{Since potential at mid-point is greater than}\\ \text{potential at infinity}(v_{\infty }=0),\text{the movement}\\ \text{of charge is against its natural tendency.}\\ \text{This gives hint that work has to be done by external force.}\end{array}$