Electric potential at any point is $V = -5x + 3y + \sqrt{15} z$,then the magnitude of the electric field is

The electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2$. Find the electric force in $N$ acting on a point charge of $2 \ C$ placed at the origin.

The electric field in a region surrounding the origin is uniform and along the $x$-axis. $A$ small circle is drawn with the centre at the origin,cutting the axes at points $A, B, C, D$ having coordinates $(a, 0), (0, a), (-a, 0), (0, -a)$ respectively,as shown in the figure. Then,the potential is minimum at the point:

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:

In a region,the potential is represented by $V(x, y, z) = 6x - 8xy - 8y + 6yz$,where $V$ is in volts and $x, y, z$ are in metres. The electric force experienced by a charge of $2 \ C$ situated at point $(1, 1, 1)$ is

For a given electric field $\vec{E} = 2x\hat{i} + 3y\hat{j}$,find the potential at $(X, Y)$ if the potential at the origin is $5\, V$.