Four positive point charges $+q$ are kept at the four corners of a square of side $l$. The net electric field at the midpoint of any one side of the square is (take $\frac{1}{4 \pi \epsilon_0}=k$ ).

The charge distribution along the semi-circular arc is non-uniform. The charge per unit length $\lambda$ is given as $\lambda = \lambda_0 \sin \theta$,with $\theta$ measured as shown in the figure. $\lambda_0$ is a positive constant. The radius of the arc is $R$. The electric field at the center $P$ of the semi-circular arc is $E_1$. The value of $\frac{\lambda_0}{\epsilon_0 E_1 R}$ is

The linear charge densities of two infinitely long thin and parallel wires are $4 \ Cm^{-1}$ and $8 \ Cm^{-1}$,and the separation between them is $4 \ cm$. The electric field intensity at the midpoint on the line joining them is:

Two charges $+4e$ and $+e$ are at a distance $x$ apart. At what distance must a charge $q$ be placed from charge $+e$ so that it is in equilibrium?

$A$ charged water drop of radius $0.1 \, \mu m$ is in equilibrium under an electric field. The charge on the drop is equivalent to the electronic charge. The intensity of the electric field is ........ $N/C$.

$A$ negative charge $-q$ is placed at the midpoint between two fixed equal positive charges $Q$, separated by a distance $2d$. If the negative charge is given a small displacement $x$ $(x \ll d)$ perpendicular to the line joining the positive charges, how will the net force $(F)$ acting on it approximately depend on $x$?