The electric field in a region of space is given as $E = (5x) \hat{i} \text{ N/C}$. Consider point $A$ on the $Y$-axis at $y = 5 \text{ m}$ and point $B$ on the $X$-axis at $x = 2 \text{ m}$. If the potentials at points $A$ and $B$ are $V_A$ and $V_B$ respectively, then $(V_B - V_A)$ is (in $\text{ V}$)

The electric potential at any point $(x, y, z)$ (all in meters) in space is given by $V = 5x^2$ volt. The electric field at the point $(1, 2, 3) \text{ m}$ is $\overrightarrow{E} = $ . . . . . . $\text{N/C}$.

The electrostatic potential inside a charged sphere is given as $V = A r^2 + B$,where $r$ is the distance from the centre of the sphere,$A$ and $B$ are constants. Then,the charge density in the sphere is

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

Electric field in a region is given by $\vec{E} = Ax\hat{i} + By\hat{j}$,where $A = 10 \ V/m^2$ and $B = 5 \ V/m^2$. If the electric potential at a point $(10, 20)$ is $500 \ V$,then the electric potential at the origin is . . . . . . $V$.