Question

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

• A. $-4\text{V}$
• B. $4\text{V}$
• C. zero
• D. $6.4\text{V}$

Answer 1

The correct option is B $4\text{V}$

Given,
Workdone in moving electron from $\text{A}$ to $\text{B}$ is, $W=6.4\times {10}^{-19}\text{J}$
charge of an electron, $q=1.6\times {10}^{-19}\text{C}$,

Now workdone in moving electron from $\text{A}$ to $\text{B}$ is,
$W=q(V_A-V_B)$

Substituting the values, we get

$6.4\times {10}^{-19}=-1.6\times {10}^{-19}\times (V_A-V_B)$

$\Rightarrow V_A-V_B=-4\text{volts}...\left(1\right)$

For the given equipotential surface,
$V_B=V_C...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow V_C-V_A=4\text{volts}$

Hence, option (c) is the correct answer.