Answer
$V = 7.39 \cdot10^{-3} \text{ V}$
Work Step by Step
The potential can be found from:
\[ V = k\int_{0}^{L} \frac{\,dq}{r}\]
Since $r=d+x$ and $dq = \lambda dx$:
\[ V = k\int_{0}^{L} \frac{\,dq}{r} = k\int_{0}^{L} \frac{\lambda dx}{d+x} = k \lambda \int_{0}^{L} \frac{1}{d+x} dx = k \lambda \left[ \ln (\mid x+d \mid) \right]_{0}^{L} \]
Since $x+d$ is positive:
\[ V = k \lambda \left[ \ln (L+d)- \ln(d) \right]\]
Since $\lambda = \cfrac{q}{L} = 4.675\cdot 10^{-13} \text{ C/m}$;
Putting the values:
\[ V = 4.675\cdot 10^{-13} \times 8.99 \cdot 10^9 \times [\ln(0.145) - \ln(0.025)] = \boxed{7.39\cdot10^{-3} \text{ V}}\]
