Topic: Electromagnetics
For a homogeneous and isotropic medium with a charge density $\rho$ and dielectric constant $\epsilon$ , the Poisson's equation for electrical potential V is
- A. $\nabla^{2}V^{\prime}=\frac{\rho}{\epsilon}$
- B. $\nabla^{2}V^{\prime}=-\frac{\rho}{\epsilon}$
- C. $\nabla\cdot V=\frac{\rho}{\epsilon}$
- D. $\nabla\cdot V=-\frac{\rho}{\epsilon}$
Answer: B. $\nabla^{2}V^{\prime}=-\frac{\rho}{\epsilon}$
Explanation: From Gauss law $\nabla\cdot\vec{D}=\rho_{v}$, $\vec{D}=\epsilon\vec{E}$. So, $\nabla\cdot(\epsilon\vec{E})=\rho_{v}$. Since medium is homogeneous and isotropic, so $\nabla\cdot\vec{E}=\frac{\rho_{v}}{\epsilon}$. Also, $\vec{E}=-\frac{\partial V}{\partial r}=-\nabla V$. Put in equation (i) $\nabla\cdot(-\nabla V)=\frac{\rho_{v}}{\epsilon} \Rightarrow \nabla^{2}V=-\frac{\rho_{v}}{\epsilon}$.