JEE Main 28 Jan 2025 Shift 1 Paper

Potential due to an infinite wire is V=2kλ‌ln‌r, where r is the distance from the wire.
Taking the point in space P(x,y,z)
Distance from wire along x-axis is rx=√y2+z2
Distance from wire along y-axis is ry=√x2+z2
Distance from wire along z-axis is rz=√x2+y2
⇒ Potential at P due to wire along x-axis is
Vx=2kλ‌ln‌rx
Potential at P due to wire along y-axis is
Vy=2kλ‌ln‌ry
Potential at P due to wire along z-axis is
Vz=2kλ‌ln‌rz
⇒ Not potential at P=V=Vx+Vy+Vz
or V=2kλ‌ln‌rx+2kλ‌ln‌ry+2kλ‌ln‌rz
i.e. V=2kλ‌ln(rxryrz)
or
‌V=2kλ‌ln(√y2+z2√z2+x2√x2+y2)
‌=kλ‌ln(y2+z2)(z2+x2)(x2+y2)
⇒ For equipotential surface
(x2+y2)(y2+z2)(z2+x2)=‌ constant ‌