Electric potential in a region is given by $\phi(x, y, z) = \phi_0 \frac{x_0}{x}$; where $x_0 = 5 \ m$ and $\phi_0 = 8 \ V$. Find the electric field at $(10 \ m, 5 \ m, 5 \ m)$.

The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (4x^2 + 8x - 3) \ V$. The value of the electric field at $x = 0.5 \ m$,in $V/m$,is:

In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$,the potential at a point $(10 \ m, 20 \ m)$ is zero. Then the potential at the origin is $\left[A=10 \ Vm^{-2}\right]$. (in $V$)

Potential gradient is defined as

Electric potential in a region is varying according to the relation $V = \frac{3x^2}{2} - \frac{y^2}{4}$,where $x$ and $y$ are in $m$ and $V$ is in $V$. Electric field intensity (in $N/C$) at a point $(1 \, m, 2 \, m)$ is:

$ABC$ is a right-angled triangle situated in a uniform electric field $\vec{E}$ which is in the plane of the triangle. The points $A$ and $B$ are at the same potential of $15 \, V$,while the point $C$ is at a potential of $20 \, V$. Given $AB = 3 \, cm$ and $BC = 4 \, cm$. The magnitude of the electric field is (in $S.I.$ units):