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 $\phi_{1}, \phi_{2}$ are equipotential surfaces, then the potential difference $\left(V_{C}-V_{A}\right)$ is

• A. -4 V
• B. 4 V
• C. zero
• D. 64 V

Answer 1

Answer: (B)

Work done by the field,
$W=q(-d V)=-e\left(V_{A}-V_{B}\right)$
$=e\left(V_{B}-V_{A}\right)=e\left(V_{C}-V_{A}\right)$
$\left(\because V_{B}=V_{C}\right)$
$\Rightarrow \left(V_{C}-V_{A}\right)=\frac{W}{e}$
$=\frac{6.4 \times 10^{-19}}{1.6 \times 10^{-19}}=4\, V$