Consider a solid insulating sphere of radius $R$ with charge density varying as $\rho = \rho_0r^2$  ($\rho_0$  is a constant and r is measure from centre).Consider two points $A$ and $B$ at distance $x$ and $y$ respectively ($x < R, y > R$) from the centre. If magnitudes of electric fields at points $A$ and $B$ are equal, then

The nuclear charge $(\mathrm{Ze})$ 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 centre of the nucleus as shown in figure The electric field is only along rhe radial direction.

Figure:$Image$

$1.$ The electric field at $\mathrm{r}=\mathrm{R}$ is

$(A)$ independent of a

$(B)$ directly proportional to a

$(C)$ directly proportional to $\mathrm{a}^2$

$(D)$ inversely proportional to a

$2.$ For $a=0$, the value of $d$ (maximum value of $\rho$ as shown in the figure) is

$(A)$ $\frac{3 Z e}{4 \pi R^3}$ $(B)$ $\frac{3 Z e}{\pi R^3}$ $(C)$ $\frac{4 Z e}{3 \pi R^3}$ $(D)$ $\frac{\mathrm{Ze}}{3 \pi \mathrm{R}^3}$

$3.$ The electric field within the nucleus is generally observed to be linearly dependent on $\mathrm{r}$. This implies.

$(A)$ $a=0$ $(B)$ $\mathrm{a}=\frac{\mathrm{R}}{2}$ $(C)$ $a=R$ $(D)$ $a=\frac{2 R}{3}$

Give the answer question $1,2$ and $3.$

Consider a uniform spherical charge distribution of radius $R_1$ centred at the origin $O$. In this distribution, a spherical cavity of radius $R_2$, centred at $P$ with distance $O P=a=R_1-R_2$ (see figure) is made. If the electric field inside the cavity at position $\overrightarrow{ r }$ is $\overrightarrow{ E }(\overrightarrow{ r })$, then the correct statement$(s)$ is(are) $Image$

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.

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $E$ without using Gauss’s law.

Electric field intensity at a point in between two parallel sheets with like charges of same surface charge densities $(\sigma )$ is