Question

A point charge $q$ held at the centre of a circle of radius $r$. Points $\text{B}$ and $\text{C}$ are two fixed points on the circumference of the circle and $\text{A}$ is a point outside the circle as shown. If $W_{AB}$ represents the work done by electric field in taking a charge $q_0$ from $\text{A}$ to $\text{B}$ and $W_{AC}$ represents the work done from $\text{A}$ to $\text{C}$, then

• A. $W_{AB}>W_{AC}$
• B. $W_{AB}<W_{AC}$
• C. $W_{AB}=W_{AC}\ne 0$
• D. $W_{AB}=W_{AC}=0$

Answer 1

The correct option is C $W_{AB}=W_{AC}\ne 0$

Electric potential at point $\text{B}$ and $\text{C}$ due to point charge $q$ will be same because they lie on the same circle.

$\therefore V_B=V_C=\frac{kq}{r}$

Thus, $V_B=V_C\ne 0$

Let the electric potential at point $\text{A}$ be $V_A$.

Work done by external force in moving $q_0$ from A $\to$ B is

$\Rightarrow W_{ext(A\to B)}=q_0(V_B-V_A)$

or, $W_{elec(A\to B)}=-W_{ext(A\to B)}$

$\therefore W_{elec(A\to B)}=(V_A-V_B)q_0......\left(1\right)$

Similarly, work done by external force in moving $q_0$ from $\text{A}\to \text{C}$ is,

$W_{ext(A\to C)}=q_0(V_C-V_A)$

or, $W_{elec(A\to C)}=-W_{ext(A\to C)}$

$\therefore W_{elec(A\to C)}=(V_A-V_C)q_0.......\left(2\right)$

From Eq.(1) and (2);

$W_{elec(A\to B)}=W_{elec(A\to C)}$

$\Rightarrow W_{AB}=W_{AC}$

$[\because V_A\ne 0\text{(A is not at infinity)}]$

Therefore, $W_{AB}=W_{AC}\ne 0$

$\begin{array}{c}\text{Why this question?}\\ \text{Tip: In a conservative field like electric field}\\ \text{work done by electric force and external force are}\\ \text{always equal and opposite. Provided that}\\ \text{charge is taken slowly}(\Delta K.E=0)\\ \therefore W_{elect}=-W_{ext}\end{array}$