Question

In moving from A to B along an electric field line, the electric field does $6.4\times {10}^{-19}J$ of work on an electron. If ${\varphi }_1,{\varphi }_2$ are equipotential surfaces, then the potential difference $(V_C-V_A)$ is?

• A. -4V
• B. 4V
• C. Zero
• D. 64V

Answer 1

The correct option is B 4V

Work done by the field $W=q(-dV)=-e(V_A-V_B)$
$=e(V_B-V_A)=e\left(\right(V_C-V_A\left)\right)$ $(\because (V_B=V_C\left)\right)$
$\Rightarrow (V_C-V_A)=\frac{W}{e}=\frac{6.4\times {10}^{-19}}{1.6\times {10}^{-19}}=4V$