Electric Field The electric field vector $\mathbf{E}$ is the negative of the gradient of the electrostatic potential $V$. That is, $\mathbf{E}=-\nabla V$. If $Q$ is a point charge at the origin, then the electrostatic potential $V$ at the point $P(x, y, z)$ is given by $V=\frac{k Q}{\sqrt{x^{2}+y^{2}+z^{2}}},$ where $k$ is a constant.
(a) Find the electric field vector $\mathbf{E}$ at the point $P(x, y, z)$.
(b) Use (a) to show that the magnitude of the electric field is $\|\mathbf{E}\|=\frac{k Q}{r^{2}},$ where $r$ is the distance from the origin to the point $P(x, y, z)$
(c) Verify that the electric field vector $\mathbf{E}$ is normal to the equipotential surfaces (surfaces on which the electrostatic potential $V$ is constant).