Question:

A thin rod is placed along the z-axis that is stretched from \(z=-d\) to \(z=+d\). Let \(\lambda\) be the linear density. There are two points \(P_1 = (0,0,2d)\) and \(P_2 = (x,0,0)\). Find the \(x\) coordinate of \(P_2\) so that potentials at \(P_1\) and \(P_2\) are equal.

Answer:

The potential at \(P_1\) is positioned at the distance of \(d\) from the edge of the rod, so the distance from the tiny piece of the charged rod should be \(2d-z\). The charge is expressed by \(\lambda dz\). Thus, the potential can be calculated as follows:

\[  \phi_1 = \int \frac{kdq}{r} = k\int^{d}_{-d}\frac{\lambda dz}{2d-z}  \]

\[  = -k\lambda \left[\ln(2d-z)\right]^{d}_{-d}  \]

\[  = -k\lambda (\ln(2d-d)-\ln(2d+d))  \]

\[  = -k\lambda (\ln(d)-\ln(3d))  \]

\[  = -k\lambda \left(\ln\frac{1}{3}\right)  \]

\[  = -k\lambda (-\ln(3)) \]

\[  = k\lambda \ln(3)  \]

Likewise,

\[  \phi_2 = k\int^{d}_{-d}\frac{\lambda dz}{\sqrt{x^2+z^2}}  \]

\[  = k\lambda \ln\frac{\sqrt{x^2+d^2}+d}{\sqrt{x^2+d^2}-d}  \]

These potentials are equal, so

\[  k\lambda \ln(3) = k\lambda \ln\frac{\sqrt{x^2+d^2}+d}{\sqrt{x^2+d^2}-d}  \]

\[  3 = \frac{\sqrt{x^2+d^2}+d}{\sqrt{x^2+d^2}-d}  \]

\[  3\sqrt{x^2+d^2}-3d = \sqrt{x^2+d^2}+d   \]

\[  2\sqrt{x^2+d^2} = 4d  \]

\[  x^2 + d^2 = 4d^2  \]

\[  x = \sqrt{3}d  \]