JEE Physics 2025 Question with Solution

Given: Three infinitely long wires with linear charge density $\lambda$ are placed along the $x$-axis, $y$-axis and $z$-axis.

Find: The equation of the equipotential surface.

For an infinite line charge, the potential varies as

$$V \propto -\ln r$$

where $r$ is the perpendicular distance from the wire.

For a point $(x,y,z)$:

• distance from the wire along the $x$-axis is $\sqrt{y^2+z^2}$,
• distance from the wire along the $y$-axis is $\sqrt{z^2+x^2}$,
• distance from the wire along the $z$-axis is $\sqrt{x^2+y^2}$.

Hence the total potential is

$$V \propto -\ln(\sqrt{y^2+z^2})-\ln(\sqrt{z^2+x^2})-\ln(\sqrt{x^2+y^2})$$

Using logarithm properties,

$$V \propto -\ln\left[\sqrt{(x^2+y^2)(y^2+z^2)(z^2+x^2)}\right]$$

For an equipotential surface, $V = \text{constant}$. Therefore,

$$(x^2+y^2)(y^2+z^2)(z^2+x^2)=\text{constant}$$

Therefore, the correct option is D.

The extracted the solution explicitly marks The Correct Option is C, but the worked steps in both approaches derive

$$(x^2+y^2)(y^2+z^2)(z^2+x^2)=\text{constant}$$

This matches option D, not option C.

So the solution working is taken, and the answer is resolved as D.