Question

If the electrostatic potential is given by $\varphi ={\varphi }_0(x^2+y^2+z^2)$ where ${\varphi }_0$ is constant, then the charge density of the given potential would be :

• A. $0$
• B. $-6{\varphi }_0{\epsilon }_0$
• C. $-2{\varphi }_0{\epsilon }_0$
• D. $\frac{-6{\varphi }_0}{{\epsilon }_0}$

Answer 1

The correct option is B $-6{\varphi }_0{\epsilon }_0$

$\overrightarrow{E}=-▽\varphi$
$\overrightarrow{▽}.\overrightarrow{E}=\rho /{ϵ}_0$
Now, $\varphi ={\varphi }_0(x^2+y^2+z^2)\Rightarrow \overrightarrow{E}=-2{\varphi }_0(\hat{i}+\hat{j}+\hat{k})\Rightarrow \rho =-6{\varphi }_0{ϵ}_0$