$A$ simple pendulum consists of a sphere with a mass of $8 \ \mu g$ and a charge of $39.2 \times 10^{-10} \ C$. When a horizontal electric field of $20 \times 10^3 \ V/m$ is applied,what angle (in degrees) does the string make with the vertical?

The coordinates of $A, B, C$ and $D$ are $(a, b, 0), (2a, 0, 0), (a, -b, 0)$ and $(0, 0, 0)$ respectively. How much work is done in moving a charge $q$ from $A$ to $D$ in a uniform electric field $\vec{E}$ directed along the positive $x$-axis?

$A$ particle with charge $q$ approaches a fixed particle with charge $Q$ with an initial velocity $v$. It comes to a stop at a closest distance $r$ from $Q$. If the particle $q$ is given an initial velocity $2v$,the closest distance will be:

$A$ charged spherical drop of mercury is in equilibrium in a plane horizontal air capacitor and the intensity of the electric field is $6 \times 10^4 \ Vm^{-1}$. The charge on the drop is $8 \times 10^{-18} \ C$. The radius of the drop is $\left[ \rho_{air} = 1.29 \ kg/m^3, \rho_{Hg} = 13.6 \times 10^3 \ kg/m^3, g = 9.8 \ m/s^2 \right]$.

Kinetic energy of an electron accelerated in a potential difference of $100 \, V$ is

$A$ particle of mass $M$ and charge $q$, initially at rest, is accelerated by a uniform electric field $E$ through a distance $D$ and is then allowed to approach a fixed static charge $Q$ of the same sign. The distance of the closest approach of the charge $q$ will then be