The formula for the electric potential of a localized charge distribution (with the convention that $V \to 0$ as the distance from the charge distribution goes to infinity) is
$\begin{align*}V(\mathbf{r}) = \int \frac{1}{4 \pi \epsilon_0}\frac{\rho(\mathbf{r}')}{\left|\mathbf{r} - \mathbf{r}' \right|}...$
In the case of this problem $\left|\mathbf{r} - \mathbf{r}' \right|$ is always equal to $\sqrt{R^2 + x^2}$ when the $\rho(\mathbf{r}')$ is nonzero. Thus,
$\begin{align*}V(\mathbf{r}) = \frac{1}{4 \pi \epsilon_0\sqrt{R^2 + x^2}} \int \rho(\mathbf{r}') d^3 x' = \boxed{\frac{1}{4 \p...$
Therefore, answer (B) is correct.

Solution to 1996 Problem 3