$A$ charged ball $B$ hangs from a silk thread $S,$ which makes an angle $\theta$ with a large charged conducting sheet $P,$ as shown in the figure. The surface charge density $\sigma$ of the sheet is proportional to $:-$

$A$ spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

The nuclear charge $(Ze)$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho(r)$ (charge per unit volume) is dependent only on the radial distance $r$ from the center of the nucleus as shown in the figure. The electric field is only along the radial direction.
$1.$ The electric field at $r=R$ is
$(A)$ independent of $a$
$(B)$ directly proportional to $a$
$(C)$ directly proportional to $a^2$
$(D)$ inversely proportional to $a$
$2.$ For $a=0$,the value of $d$ (maximum value of $\rho$ as shown in the figure) is
$(A)$ $\frac{3Ze}{4\pi R^3}$ $(B)$ $\frac{3Ze}{\pi R^3}$ $(C)$ $\frac{4Ze}{3\pi R^3}$ $(D)$ $\frac{Ze}{3\pi R^3}$
$3.$ The electric field within the nucleus is generally observed to be linearly dependent on $r$. This implies
$(A)$ $a=0$ $(B)$ $a=\frac{R}{2}$ $(C)$ $a=R$ $(D)$ $a=\frac{2R}{3}$
Give the answer for questions $1, 2,$ and $3.$

The electric field at $20 \,cm$ from the centre of a uniformly charged non-conducting sphere of radius $10 \,cm$ is $E$. Then at a distance $5 \,cm$ from the centre it will be

Obtain Coulomb's law from Gauss's law.

An infinitely long thin straight wire has a uniform charge density of $\frac{1}{4} \times 10^{-2} \text{ C/m}$. What is the magnitude of the electric field at a distance of $20 \text{ cm}$ from the axis of the wire?