Consider a sphere of radius $R$ which carries a uniform charge density $\rho$. If a sphere of radius $\frac{R}{2}$ is carved out of it,as shown,the ratio $\frac{|\overrightarrow{E}_{A}|}{|\overrightarrow{E}_{B}|}$ of the magnitude of the electric field $\overrightarrow{E}_{A}$ and $\overrightarrow{E}_{B}$ respectively,at points $A$ and $B$ due to the remaining portion is

Gauss's law can help in the easy calculation of the electric field due to:

The electric field intensity on the surface of a solid charged sphere of radius $r$ and volume charge density $\rho$ is ( $\epsilon_0=$ permittivity of free space).

Find the electric field due to an infinitely long charged cylinder of radius $R$ having a linear charge density $\lambda$ at a distance of $R/2$ from its axis.

An infinitely long thin straight wire has a uniform linear charge density of $\frac{1}{3} \, C \cdot m^{-1}$. The magnitude of the electric field intensity at a point $18 \, cm$ away is (given $\varepsilon_0 = 8.85 \times 10^{-12} \, C^2 \cdot N^{-1} \cdot m^{-2}$):

$A$ uniform rod $AB$ of mass $m$ and length $l$ is hinged at its midpoint $C$. The left half $(AC)$ of the rod has linear charge density $-\lambda$ and the right half $(CB)$ has $+\lambda$,where $\lambda$ is constant. $A$ large non-conducting sheet of uniform surface charge density $\sigma$ is also present near the rod. Initially,the rod is kept perpendicular to the sheet. The end $A$ of the rod is initially at a distance $d$. Now,the rod is rotated by a small angle $\theta$ in the plane of the paper and released. The time period of small angular oscillations is: