Question

When a $2\mu \text{C}$ charge is carried from point $\text{A}$ to point $\text{B}$, the amount of workdone by the electric field is $50\mu \text{J}$. Find the potential difference and also find which point is at a higher potential.

• A. $25\text{V},\text{B}$
• B. $25\text{V},\text{A}$
• C. $20\text{V},\text{B}$
• D. Both are at the same potential.

Answer 1

The correct option is B $25\text{V},\text{A}$

Given,
Charge of the particle $q=2\mu \text{C}$
Workdone by the electric field $W=50\mu \text{J}$

We know, workdone by electric field on a particle of charge $q$ moving between two points $\text{A}$ and $\text{B}$ is given by

$W=q(V_A-V_B)$

$\Rightarrow V_A-V_B=\frac{W}{q}$

Substituting the given data we get,

$V_A-V_B=\frac{50\times {10}^{-6}}{2\times {10}^{-6}}=25\text{V}$

From this, we can say that $V_A>V_B$

Hence, option (b) is the correct answer.