The insulation property of air breaks down at $E = 3 \times 10^6 \ V/m$. The maximum charge that can be given to a sphere of diameter $5 \ m$ is approximately (in coulombs):

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

$A$ uniformly charged rod of length $4\,cm$ and linear charge density $\lambda = 30\,\mu C/m$ is placed as shown in the figure. Calculate the $x-$ component of the electric field at point $P$.

Two conducting spheres of radii $r_1$ and $r_2$ are charged to the same surface charge density. The ratio of electric fields near their surfaces is

$A$ wire of length $L$ has a charge $Q$ distributed uniformly along its length. The wire is bent in the shape of a semicircle. The magnitude of the electric field at the centre of curvature of the semicircle is

$A$ metal sphere of radius $R \ cm$ is charged with $4 \pi \mu C$ and is situated in air. If $\sigma$ is the surface charge density and $E$ is the electric intensity at a distance $r$ from the centre of the sphere,then $r$ is equal to ($\epsilon_{0}$ is the permittivity of free space).