Solution to 1996 Problem 3

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.

return to the 1996 problem list

return to homepage

Please send questions or comments to where X = physgre.