Answer
The electric potential at point P is $~~2.57\times 10^4~V$
Work Step by Step
Let's assume that there is a charged rod with one end at the origin with a length of $5.00D$.
We can use Equation (24-35) to find the electric potential at point P:
$V_1 = \frac{\lambda}{4\pi~\epsilon_0}~ln~[\frac{5.00D+\sqrt{(5.00D)^2+D^2}}{D}]$
$V_1 = \frac{2.00\times 10^{-6}~C/m}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~ln~(5.00+\sqrt{26.6})$
$V_1 = 4.1567\times 10^4~V$
Let's assume that there is a charged rod with one end at the origin with a length of $D$.
We can use Equation (24-35) to find the electric potential at point P:
$V_2 = \frac{\lambda}{4\pi~\epsilon_0}~ln~[\frac{D+\sqrt{D^2+D^2}}{D}]$
$V_2 = \frac{2.00\times 10^{-6}~C/m}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~ln~(1.00+\sqrt{2})$
$V_2 = 1.5843\times 10^4~V$
By superposition, the electric potential at point P due to the rod in the problem is $V = V_1-V_2$.
We can find the electric potential at point P:
$V = V_1 - V_2$
$V = 4.1567\times 10^4~V-1.5843\times 10^4~V$
$V = 2.57\times 10^4~V$
The electric potential at point P is $~~2.57\times 10^4~V$.
