If the electrostatic potential were given by $\phi=\phi_{0}\left(x^{2}+y^{2}+z^{2}\right)$, where $\phi_{0}$ is constant, then the charge density giving rise to the above potential would be.

Question

Aicharya · Accepted Answer

$\begin{aligned} E &=-
abla \phi=-\phi_{0} 2[x \hat{i}+y \hat{i}+z \hat{x}] \ &=\varepsilon_{0} 
abla \cdot E=-2 \varepsilon_{0} \phi_{0} 
abla \cdot(x \hat{i}+y \hat{i}+z \hat{x}) \ n &=-6 \phi_{0} \varepsilon_{0} \end{aligned}$ ^

If the electrostatic potential were given by $\phi=\phi_{0}\left(x^{2}+y^{2}+z^{2}\right)$, where $\phi_{0}$…

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