Question

FIGURE P25.69 shows a thin rod of length L and charge Q. Find an expression for the electric potential a distance x away from the center of the rod on the axis of the rod.

Short answer

The electric potential is given as$V=\frac{kQ}{L}In\frac{2x+L}{2x-L}$

Step-by-step solution

Given Information

We need to find an expression for the electric potential a distance$x$ away from the center of the rod on the axis of the rod.

Simplify

Consider the fact that the potential is a scalar quantity, which means

$V=k\frac{q}{r}$

We can express this as an integral. In our case, instead of the charge, we'll have the linear charge density $Q/L$, and the integration variable, let it be$l$. The potential at point $x$will therefore be given as

$V=k{}_{\raisebox{1ex}{$-L$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{}^{\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{Q/L}{l+x}dl$

Solving this integral,

$V=\frac{kQ}{L}{}_{-L/2}{}^{L/2}\frac{dl}{l+x}=\frac{kQ}{L}{\left[\ln \left(l+x\right)\right]}_{-L/2}^{L/2}=\frac{kQ}{L}\left[\ln \left(x+L/2\right)-\ln (x_L/2\right]$

By the properties of the logarithm and by simplifying with the factor of two, we get

$V=\frac{kQ}{L}In\frac{2x+L}{2x-L}$