The figure shows two equipotential lines in the $x, y$ plane for an electric field. The scales are marked in $\text{cm}$. The $x$-component $E_x$ and $y$-component $E_y$ of the electric field in the space between these equipotential lines are respectively:

The electric potential $V(x, y, z)$ for a planar charge distribution is given by:
$V(x, y, z) = \begin{cases} 0 & \text{for } x < -d \\ -V_0(1 + \frac{x}{d})^2 & \text{for } -d \le x < 0 \\ -V_0(1 + 2\frac{x}{d}) & \text{for } 0 \le x < d \\ -3V_0 & \text{for } x \ge d \end{cases}$
where $-V_0$ is the potential at the origin and $d$ is a distance. The graph of the electric field as a function of position is:

In a certain region,the electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \, V$. The magnitude of the force on a charge of $2 \, C$ at the origin is ........ $N$.

Obtain the relation between electric field and electric potential.

Two parallel plates are separated by a distance of $5 \, mm$. The potential difference between them is $50 \, V$. $A$ particle of mass $10^{-15} \, kg$ and charge $10^{-11} \, C$ enters the region with a velocity of $10^7 \, m/s$. The acceleration of the particle will be:

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?