JEE Main 2025 — Electrostatics Question with Solution

Electric field due to infinitely long wire

$$\overrightarrow{E}=-\frac{2k\lambda }{r}\hat{r}$$

We know, $$V=-\int \overrightarrow{E}.\overrightarrow{dr}$$

$$=-\int \frac{-2k\lambda }{r}\hat{r}.\overrightarrow{dr}$$

$$v=\int \frac{2k\lambda }{r}dr=2k\lambda \ln r+c$$

Net potential due to all wires,

$$V=v_1+v_2+v_3$$

$$=2k\lambda \ln \left(\sqrt{x^2+y^2}\right)+2k\lambda \ln \left(\sqrt{y^2+z^2}\right)+2k\lambda \ln \left(\sqrt{z^2+x^2}\right)+c$$

for equipotential surface, $$v=c$$

$$\Rightarrow 2k\lambda \left[\ln \left(\sqrt{x^2+y^2}.\sqrt{y^2+z^2}.\sqrt{z^2+x^2}\right)\right]=c$$

$$\Rightarrow \sqrt{(x^2+y^2).(y^2+z^2).(z^2+x^2)}=c$$

$$\Rightarrow (x^2+y^2)(y^2+z^2)(z^2+x^2)=c$$

where, c = constant.