Question

As shown in the figure, on bringing a charge $Q$ from point $\text{A}$ to $\text{B}$ and from $\text{B}$ to $\text{C}$, the work done are $2\text{Joules}$ and $-3\text{Joules}$ respectively. The work done in bringing the charge from $\text{C}$ to $\text{A}$ will be

• A. $-1\text{Joule}$
• B. $1\text{Joule}$
• C. $2\text{Joules}$
• D. $5\text{Joules}$

Answer 1

The correct option is B $1\text{Joule}$

Given:
$W_{A\to B}=2\text{Joules};W_{B\to C}=-3\text{Joules}$

Since electric field is a conservative field, so net work done in a loop will be zero.

$\therefore W_{A\to B}+W_{B\to C}+W_{C\to A}=0$

$\Rightarrow W_{C\to A}=-(W_{A\to B}+W_{B\to C})$

$\Rightarrow W_{C\to A}=-[2+(-3\left)\right]=1\text{Joule}$

Hence, option $\left(b\right)$ is the correct answer.

Why this question: Exploiting the knowledge of workdone in a conservative force field.