The electrostatic potential in a charged spherical region of radius $r$ varies as $V = ar^3 + b$, where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $\alpha \times \pi a \epsilon_0$. The value of $\alpha$ is . . . . . . .

Electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2$. Then,the electric force acting on a $2 \, C$ point charge placed at the origin will be......$N$.

The potential is varying with distance $(x, y)$ as $V = \frac{1}{2} (y^2 - 4x) \text{ V}$. The electric field at $x = 1 \text{ m}$ and $y = 1 \text{ m}$ 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):

As shown in the figure, if the values of the electric potential at three points $A, B$ and $C$ in a uniform electric field $(\vec{E})$ are $V_A, V_B$, and $V_C$ respectively, then

Two plates are $20 \ cm$ apart and a potential difference of $10 \ V$ is applied between them. The electric field between the plates is . . . . . . . (in $Vm^{-1}$)