The work done in carrying a charge of $5\,\mu \,C$ from a point $A$ to a point $B$ in an electric field is $10\,mJ$. The potential difference $({V_B} - {V_A})$ is then
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(a) Work done $W = Q({V_B} - {V_A}) \Rightarrow \,({V_B} - {V_A}) = \frac{W}{Q}$ $ = \frac{{10 \times {{10}^{ - 3}}}}{{5 \times {{10}^{ - 6}}}}J/C = 2\,kV$
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