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 $4V$

Electric field $=6.4\times {10}^{-19}J$

work on electron $q_1$ and ${\varphi }_2$ potential difference

$V_C-V_A=\frac{w}{q}=\frac{6.4\times {10}^{-19}}{1.6\times {10}^{-19}}=4V$

$\because$ $q=1.6\times {10}^{-19}C$ due to electron movement.