JEE Physics 2026 Question with Solution

Given: $q_1=1\,\text{nC}$, $q_2=2\,\text{nC}$, charge brought $q=3\,\text{nC}$, side of equilateral triangle $r=3\,\text{cm}=0.03\,\text{m}$.

Find: The work done in bringing the charge to the third corner.

Concept: The work done in bringing a charge from infinity to a point in an electric field is

$$W=qV$$

where $V$ is the electric potential at that point due to the existing charges.

Electric potential due to a point charge is

$$V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}$$

Since the triangle is equilateral, the third corner is at the same distance from both charges. Therefore,

$$V=\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r}+\frac{q_2}{r}\right)=\frac{1}{4\pi\varepsilon_0}\frac{q_1+q_2}{r}$$

Substitute the values:

$$V=9\times10^9\times\frac{(1+2)\times10^{-9}}{0.03}$$ $$V=9\times10^9\times\frac{3\times10^{-9}}{0.03}=\frac{27}{0.03}=900\,\text{V}$$

Now calculate the work done:

$$W=qV=3\times10^{-9}\times900$$ $$W=2700\times10^{-9}\,\text{J}=2.7\times10^{-6}\,\text{J}$$

Thus,

$$W=2.7\,\mu\text{J}$$

Therefore, the correct option is D.

Given: Two charges are already fixed at two corners, and the third charge is brought from infinity.

Find: The external work required.

Instead of finding force at each stage, first compute the electric potential at the third corner due to the two fixed charges. Because potential is a scalar quantity, the contributions add directly.

1. Potential due to $q_1$ at the third corner:

$$V_1=\frac{1}{4\pi\varepsilon_0}\frac{q_1}{r}$$

2. Potential due to $q_2$ at the third corner:

$$V_2=\frac{1}{4\pi\varepsilon_0}\frac{q_2}{r}$$

3. Total potential:

$$V=V_1+V_2=\frac{1}{4\pi\varepsilon_0}\frac{q_1+q_2}{r}$$

4. Substitute:

$$V=9\times10^9\times\frac{3\times10^{-9}}{0.03}=900\,\text{V}$$

5. Work done in bringing $q=3\,\text{nC}$:

$$W=qV=3\times10^{-9}\times900=2.7\times10^{-6}\,\text{J}$$

So the required work is $2.7\,\mu\text{J}$, hence option D.