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:

If the potential energy of an electron and a proton at a distance $r$ is given by the formula $U = - \left( \frac{ke^2}{3r^3} \right)$,then which law of force applies?

The electric potentials of two plates are $-10\, V$ and $+30\, V$ respectively. If the distance between the two plates is $2\, cm$,then the electric field between them is ... $V/m$.

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.

When a charge of $3 \, C$ is placed in a uniform electric field, it experiences a force of $3000 \, N$. Within this field, the potential difference between two points separated by a distance of $1 \, cm$ is: (in $V$)