A spherically symmetric charge distribution is considered with charge density varying as

$\rho(r)=\left\{\begin{array}{ll}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { Zero } & \text { for } r>R\end{array}\right.$

Where, $r ( r < R )$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be.

An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho$. It has a spherical cavity of radius $R / 2$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$, which is at a distance $2 \ R$ from the axis of the cylinder, is given by the expression $\frac{23 \rho R }{16 k \varepsilon_0}$. The value of $k$ is

A spherical conductor of radius $10\, cm$ has a charge of $3.2 \times 10^{-7} \,C$ distributed uniformly. What is the magnitude of electric field at a point $15 \,cm$ from the centre of the sphere?

$\left(\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} Nm ^{2} / C ^{2}\right)$

Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho(r) = kr$, where $r$ is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance, $a$, from the centre. If $A$ and $B$ do not experience any force, then

The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero ?

The volume charge density of a sphere of radius $6 \,m$ is $2 \,\mu cm ^{-3}$. The number of lines of force per unit surface area coming out from the surface of the sphere is $....\times 10^{10}\, NC ^{-1}$.  [Given : Permittivity of vacuum  $\left.\epsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1}- m ^{-2}\right]$