cse โข paper_1 โข electrodynamics
[UPSC CSE 2005] Electrostatics and Magnetostatics (Q200, 20M)
(a) Prove the relation \nabla^2 \left( \frac{1}{|\vec{r} - \vec{r}'|} \right) = - 4 \pi \delta \left(|\vec{r} - \vec{r}'|\right) and hence show that \phi(\vec{r}) = \int \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|} \, d\vec{r}' is a solution of the Poisson equation \nabla^2 \phi(\vec{r}) = - 4 \pi \rho(\vec{r}).
Discussion (0)
Sign in to post a solution, derivation, or discussion.
No comments yet. Start the conversation!