If the electric potential in a region is given by $V = 4x^2$ volts,then the electric field at the point $(1, 0, 2) \ m$ is:

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 . . . . . . .

Some equipotential surfaces are shown in the figure below. The magnitude of the electric field will be $:-$ (in $V/m$)

The electrostatic potential inside a charged spherical ball is given by $V = ar^2 + b$,where $r$ is the distance from its centre and $a$ and $b$ are constants. The volume charge density of the ball is [$\varepsilon_0$ = permittivity of free space].

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

For a charged spherical ball,electrostatic potential inside the ball varies with $r$ as $V = 2ar^2 + b$. Here,$a$ and $b$ are constants and $r$ is the distance from the center. The volume charge density inside the ball is $-\lambda a \varepsilon$. The value of $\lambda$ is $...........$. $\varepsilon =$ permittivity of medium.