Two concentric rings, one of radius $a$and the other of radius $b$, have the charges $+q$and $-{\left(\frac{2}{5}\right)}^{-\frac{3}{2}}q$, respectively, as shown in the figure.

Find the ratio $\frac{b}{a}$ if a charged particle placed on the axis at $z=a$is in equilibrium.

The magnitude of electric field intensity $E$ is such that, an electron placed in it would experience an electrical force equal to its weight, is given by

Four point charges $q_A=2 \mu C, q_B=-5 \mu C, q_C=2 \mu C, q_D=-5 \mu C$ are located at the corners of a square $ABCD$ of side $10 \mathrm{cm}.$ The force acting on a charge of $1 \mu \mathrm{C}$ placed at centre of the square is

An imaginary equilateral triangle $ABC$ of side length $2 \mathrm{m}$ is placed in a uniform electric field $\overrightarrow{\mathrm{E}}=10 \mathrm{N} {\mathrm{C}}^{-1}$ as shown. Then. $V_A-V_B=$

As shown in the figure below, a point charge $q$ moves from point $P$ to a point $S$ traversing a path $PQRS$ in a uniform $\overrightarrow{E}.$ The electric field is directed along a direction parallel to $x$-axis. The coordinates of, $P,Q,R$ and $S$ are $\left(a,b,0\right)\left(2a,0,0\right)\left(a,-b,0\right)$ and $\left(0,0,0\right)$ respectively. What is the work done by the field in the process?