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 $\begin{aligned}
& V=2 k \lambda \ln \left(\sqrt{y^2+z^2} \sqrt{z^2+x^2} \sqrt{x^2+y^2}\right) \
& =k \lambda \ln \left(y^2+z^2\right)\left(z^2+x^2\right)\left(x^2+y^2\right)
\end{aligned}\RightarrowForequipotentialsurface\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=\text { constant }$
Three infinitely long wires with linear charge density λ are placed along the x-a x i s, y-axis and z- axis respectively. Which of the following…
JEE Main 2025 — Physics Electromagnetism
Three infinitely long wires with linear charge density λ are placed along the x−axis,y-axis and z− axis respectively. Which of the following denotes an equipotential surface?
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
xyz= constant
- B
xy+yz+zx= constant
- C
(x2+y2)(y2+z2)(z2+x2)= constant
- D
(x+y)(y+z)(z+x)= constant
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