JEE Main 2025 — Electrostatics Question with Solution

To determine the value of $K$ for the given electrical potential scenario, we start by calculating the potential at points C and D.

Potential at Point C:

The potential at point C, $V_C$, is given by the sum of the potentials due to charges $q_1$ and $q_2$:

$V_C=\frac{k{q}_1}{0.4}+\frac{k{q}_2}{0.5}$

Potential at Point D:

Similarly, the potential at point D, $V_D$, is:

$V_D=\frac{k{q}_1}{0.4}+\frac{k{q}_2}{0.1}$

Change in Potential Energy:

The change in potential energy $\Delta U$ when $q_3$ moves from C to D is:

$\Delta U=(V_D-V_C)\times q_3$

Substituting the expressions for $V_D$ and $V_C$, we get:

$\Delta U=\left(\frac{k{q}_2}{0.1}-\frac{k{q}_2}{0.5}\right)\times q_3$

Simplifying the Expression:

By simplifying the expression, we find:

$\Delta U=\left(\frac{5k{q}_2}{0.5}\right)\times q_3-\left(\frac{k{q}_2}{0.5}\right)\times q_3$

$\Delta U=\frac{4k{q}_2}{0.5}\times q_3=8k{q}_2{q}_3$

$\Delta U=\frac{8q_2{q}_3}{4\pi {\epsilon }_0}$

Thus, the value of $K$ is $8q_2$.