If the atmospheric electric field is approximately $150 \,V/m$ and the radius of the Earth is $6400 \,km$,then the total charge on the Earth's surface is .......... coulomb.

$A$ cube of side $a$ is placed in an electric field $\vec{E} = E_0 x \hat{i}$. What is the total charge enclosed by the cube?

Let $\rho (r) = \frac{Q}{\pi R^4} r$ be the volume charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point $p$ inside the sphere at distance $r_1$ from the centre of the sphere,the magnitude of the electric field is:

An infinite non-conducting sheet has a surface charge density $2 \times 10^{-7} \text{ C/m}^2$ on one side. The distance between two equipotential surfaces whose potential difference is $90 \text{ V}$ is (assume $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2$):

$A$ hollow sphere of charge does not produce an electric field at any

Electric field intensity at points in between and outside two thin separated parallel sheets of infinite dimension with like charges of same surface charge density $(\sigma)$ are . . . . . . and . . . . . . respectively.