Three infinitely long wires with linear charge density $\lambda$ are placed along the $x-a x i s, y$-axis and $z-$ axis respectively. Which of the following denotes an equipotential surface?

Question

Aicharya · Accepted Answer

Potential due to an infinite wire is $V=2 k \lambda \ln r$, where $r$ is the distance from the wire.
Taking the point in space $P(x, y, z)$
Distance from wire along $x$-axis is $r_x=\sqrt{y^2+z^2}$
Distance from wire along $y$-axis is $r_y=\sqrt{x^2+z^2}$
Distance from wire along $z$-axis is $r_z=\sqrt{x^2+y^2}$
$\Rightarrow$ Potential at $P$ due to wire along $x$-axis is
$V_x=2 k \lambda \ln r_x$
Potential at $P$ due to wire along $y$-axis is
$V_y=2 k \lambda \ln r_y$
Potential at $P$ due to wire along $z$-axis is
$V_z=2 k \lambda \ln r_z$
$\Rightarrow$ Not potential at $P=V=V_x+V_y+V_z$
or $\quad V=2 k \lambda \ln r_x+2 k \lambda \ln r_y+2 k \lambda \ln r_z$
i.e. $V=2 k \lambda \ln \left(r_x r_y r_z\right)$
or
$\begin{aligned}
& V=2 k \lambda \ln \left(\sqrt{y^2+z^2} \sqrt{z^2+x^2} \sqrt{x^2+y^2}\right) \\ & =k \lambda \ln \left(y^2+z^2\right)\left(z^2+x^2\right)\left(x^2+y^2\right)
\end{aligned}$
$\Rightarrow$ For equipotential surface
$\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=\text { constant }$

Three infinitely long wires with linear charge density $\lambda$ are placed along the $x-a x i s, y$-axis…

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