Physics - Elecrostatics Question with Solution | TestHub

Potential due to an infinite wire is $V=2k\lambda \ln r$, 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 $V_x=2k\lambda \ln r_x$ Potential at $P$ due to wire along $y$-axis is $V_y=2k\lambda \ln r_y$ Potential at $P$ due to wire along $z$-axis is $V_z=2k\lambda \ln r_z$ $\Rightarrow$ Not potential at $P=V=V_x+V_y+V_z$ or $V=2k\lambda \ln r_x+2k\lambda \ln r_y+2k\lambda \ln r_z$ i.e. $V=2k\lambda \ln \left(r_x{r}_y{r}_z\right)$ or $\begin{array}{rl}& V=2k\lambda \ln \left(\sqrt{y^2+z^2}\sqrt{z^2+x^2}\sqrt{x^2+y^2}\right)\\ & =k\lambda \ln \left(y^2+z^2\right)\left(z^2+x^2\right)\left(x^2+y^2\right)\end{array}$ $\Rightarrow$ For equipotential surface $\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=\text{ constant }$