Find out the electric field intensity at point $A(1, 0, 2)$ due to a point charge $-20\,\mu C$ situated at point $B(0, 2, 1)$.

$A$ thin metallic wire having a cross-sectional area of $10^{-4} \, m^2$ is used to make a ring of radius $30 \, cm$. $A$ positive charge of $2 \pi \, pC$ is uniformly distributed over the ring, while another positive charge of $30 \, pC$ is kept at the centre of the ring. The tension in the ring is . . . . . . $N$; provided that the ring does not get deformed (neglect the influence of gravity). (Given, $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, SI$ units)

The potential in an electric field varies as $V = (x^2 - y^2)$. The electric lines of force in the $X-Y$ plane are

Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q.$ The distance of the point $P$ from the centre of the rod is $a = \frac{\sqrt{3}}{2} L$.

What is the direction of electric field intensity?

$A$ 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,