Question

The figure shows a set of equipotential surfaces. There are a few points marked on them. An electron is being moved from one point to other. Which of the following statement is/are correct?

• A. As the electron moves from $E$ to $A$, the potential energy increases.
• B. Work done by the electric field, in moving the electron from $C$ to $D$ is same as from $D$ to $E$.
• C. The electric field is directed along +$x$ -axis.
• D. Work done by the electric field, in moving the electron from $B$ to $C$, is positive.

Answer 1

Answer: (A), (B), (D)

Work done by the electric field, in moving the electron from $B$ to $C$, is positive.

Electric field lines are always directed from region of high potential to low potential and the $\overrightarrow{E}$ is perpendicular to the equipotential surfaces.
$\Rightarrow \overrightarrow{E}$ is along negative $x$- axis
$\therefore$ option (a) is incorrect


Electron being a negative charged particle, it will experience force in a direction opposite to $\overrightarrow{E}$ i.e along the positive $x$-axis.

Thus, electrostatic force ($\overrightarrow{F}$) and displacement $\left(\overrightarrow{S}\right)$ are in same direction when electron is displaced from $B$ to $C$, hence work done ($W$) by electrostatic force on an electron is positive.
$\therefore$ Option (b) is the correct answer.

Work done by electric force is, $W=-q(V_f-V_i)=-q\Delta V$

The change in potential when electron moves from $C$ to $D=\Delta V_{CD}=5\text{V}$

The change in potential when electron moves from $D$ to $E=\Delta V_{DE}=5\text{V}$

Charge of electron and $\Delta V$ is constant for both the cases,
$\therefore W_{CD}=W_{DE}$

Also, electron is moving in the same direction of the electric field hence its potential energy will increase. So, options (c) and (d) are also correct.

Hence, options (b) , (c) and (d) are the correct alternatives.

Why this question ? Tip : whenever work is done against a conservative field, the associated potential energy of body will increase.