Question

In moving from $A$ to $B$ along an electric field line, the electric field does $1.28\times {10}^{-18}J$ of work on an electron. If ${\varphi }_1,{\varphi }_2$ are equipotential surface, then potential difference $V_E-V_D$ is equal to :

• A. $-8$ volt
• B. $+8$ volt
• C. zero
• D. $64$ volt

Answer 1

The correct option is A $-8$ volt

Work done by field $W=-\Delta U=U_A-U_B=(-e)V_A-(-e)V_B$

$\Rightarrow$ $1.28\times {10}^{-18}=1.6\times {!0}^{-19}(V_B-V_A)$
$V_B-V_A=8$ volt $\Rightarrow$ $V_E-V_D=V_A-V_B=-8$volt