Question

When a particle carrying a charge of $10\times {10}^{-6}\mathrm{coulombs}$ is brought from infinity to a point P, $2\times {10}^{-3}\mathrm{joules}$ of work is done. What is the potential at the point P?

Answer 1

Concept used: Potential difference

Prerequisite: Knowledge of potential difference
Step 1: Given data,

1. The charge on particle is $q=10\times {10}^{-6}\mathrm{C}$.
2. The work done to move the charge is $W=2\times {10}^{-3}\mathrm{J}$.

Step 2: Calculate the potential,

Considering the work done at infinity is zero, i.e., $V_{\infty }=0$.

The potential at any point due to the movement of the charge is given by,

$\begin{array}{rcl}∆V& =& \frac{W}{q}\\ V_P-V_{\infty }& =& \frac{W}{q}\end{array}$

Substitute the given values in above equation,

$\begin{array}{rcl}{V}_P-0& =& \frac{2\times {10}^{-3}\mathrm{J}}{10\times {10}^{-6}\mathrm{C}}\\ V_P& =& 200\mathrm{V}\end{array}$

Thus, the potential at point P is 200 V.