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 $V$ at any point $(x, y, z)$ in space is given by the equation $V = 4x^2 \, V$,where $x, y$,and $z$ are all in meters. The electric field at the point $(1 \, m, 0, 2 \, m)$ in $V/m$ 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:

The figure shows the variation of electric field intensity $E$ versus distance $x$. What is the potential difference between the points at $x = 2 \, m$ and $x = 6 \, m$ from $O$ (in $V$)?

Find the potential $V$ of an electrostatic field $\vec{E} = a(y\hat{i} + x\hat{j})$,where $a$ is a constant.

Two metal plates are separated by $2 \,cm$. The potentials of the plates are $-10 \,V$ and $+30 \,V$. The electric field between the two plates is: (in $\,V/m$)