Answer
The electric potential at point P is $~~0.0243~V$
Work Step by Step
We can use Equation (24-35) to find the electric potential at point P due to the left half of the rod:
$V = \frac{\lambda}{4\pi~\epsilon_0}~ln~[\frac{L/2+\sqrt{(L/2)^2+d^2}}{d}]$
$V = \frac{3.68\times 10^{-12}~C/m}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~ln~[\frac{(0.030~m)+\sqrt{(0.030~m)^2+(0.080~m)^2}}{0.080~m}]$
$V = 0.012129~V$
By symmetry, the electric potential due to the right half of the rod is also equal to this value.
We can find $V_{net}$:
$V_{net} = (2)(0.012129~V)$
$V_{net} = 0.0243~V$
The electric potential at point P is $~~0.0243~V$.
