An infinite non-conducting sheet has a surface charge density of $7 \times 10^{-7} \text{ C m}^{-2}$ on one side. The distance between equipotential surfaces whose potentials differ by $19.8 \text{ V}$ will be (Take $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^9 \text{ SI units}$) (in $\text{ mm}$)

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:

Find the potential $V$ of an electrostatic field $\vec{E} = a(y\hat{i} + x\hat{j})$,where $a$ is a constant.

Two plates are separated by a distance of $20 \, cm$. The potential difference between them is $10 \, V$. The electric field between the two plates is ...... $V m^{-1}$.

Assume that an electric field $E=20 x^2 \hat{i}$ exists in space. If $V_0$ is the potential at the origin and $V_A$ is the potential at $x=3 \ m$,then the potential difference $V_A-V_0$ in volts is

The potential difference between two parallel plates is $10^4 \,V$. If the plates are separated by $0.5 \,cm$, the force on an electron between the plates is