Two charges, each equal to $-q$, are kept at $(-a, 0)$ and $(a, 0)$. $A$ charge $q$ is placed at the origin. If $q$ is given a small displacement $y$ along the $y$-direction, the force acting on $q$ is proportional to:

If the net electric field at point $P$ along the $Y$-axis is zero,then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5 \sqrt{x}}$,where $x = . . . . . .$

$A$ thin glass rod is bent in a semicircle of radius $R$. $A$ charge is non-uniformly distributed along the rod with a linear charge density $\lambda = \lambda_0 \sin \theta$ (where $\lambda_0$ is a positive constant and $\theta$ is the angle with the $x$-axis). The electric field at the centre $P$ of the semicircle is,

$A$ 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 ........ $N/C$.

An electric field is given by $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$. $A$ charge $Q$ is moved by a displacement vector $\vec{r} = a \hat{i} + b \hat{j}$. The work done is:

The charges $2q, -q, -q$ are located at the vertices of an equilateral triangle. At the center of the triangle: