$A$ charge $Q$ is uniformly distributed over a large square plate of copper. The electric field at a point very close to the centre of the plane is $10 \ V/m$. If the copper plate is replaced by a plastic plate of the same geometrical dimensions and carrying the same charge $Q$ uniformly distributed,then the electric field at the point $P$ will be......$V/m$.

Explain by graph how the electric field by a thin spherical shell depends on the distance of a point from the centre.

Obtain the expression for the electric field due to a uniformly charged spherical shell at a point outside it.

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 $3.$

Within a spherical charge distribution of charge density $\rho(r)$,$N$ equipotential surfaces of potential $V_0, V_0 + \Delta V, V_0 + 2\Delta V, \dots, V_0 + N\Delta V$ (where $\Delta V > 0$) are drawn and have increasing radii $r_0, r_1, r_2, \dots, r_N$,respectively. If the difference in the radii of the surfaces is constant for all values of $V_0$ and $\Delta V$,then:

An infinite line charge produces a field of $9 \times 10^4 \ NC^{-1}$ at a distance of $2 \ cm$. The linear charge density is