Electrostatics Part 3

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 $r_z=\sqrt{x^2+y^2}$⇒ 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$V_z=2 k \lambda \ln r_z$⇒ Not potential at $P=V=V_x+V_y+V_z$or $V=2 k \lambda \ln r_x+2 k \lambda \ln r_y+2 k \lambda \ln r_z$i.e. $V=2 k \lambda \ln (r_x r_y r_z)$or$\;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 (y^2+z^2) (z^2+x^2) (x^2+y^2)$⇒ For equipotential surface$(x^2+y^2) (y^2+z^2) (z^2+x^2) =\; \text{ constant } \;$