The electric potential at a point $(x, y, z)$ is given by $V = -x^2y - xz^3 + 4$. The electric field at that point is:

The electric potential inside a charged sphere is given by $\phi = ar^2 + b$,where $r$ is the distance from the center,and $a$ and $b$ are constants. The charge density inside the sphere is .......

If the electric potential is given by $V = 4x + 3y$,what is the electric field at the point $(2, 1)$?

The potential difference between two parallel plates is $10^4 \,V$. If the plates are separated by $0.5 \,cm$, the force on an electron between the plates is

The electric potential $V(x, y, z)$ for a planar charge distribution is given by:
$V(x, y, z) = \begin{cases} 0 & \text{for } x < -d \\ -V_0(1 + \frac{x}{d})^2 & \text{for } -d \le x < 0 \\ -V_0(1 + 2\frac{x}{d}) & \text{for } 0 \le x < d \\ -3V_0 & \text{for } x \ge d \end{cases}$
where $-V_0$ is the potential at the origin and $d$ is a distance. The graph of the electric field as a function of position is:

In space,the electric potential varies as $V = 20|\vec{r}|$ volt,where $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ is the position vector. Then,the electric field in $N C^{-1}$ at the point $(4 \ m, 3 \ m, -5 \ m)$ is: