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 characterised by a charge density having the following variations

$\rho (r)\, = \,{\rho _0}\left( {1 – \frac{r}{R}} \right)$ for $r < R$

$\rho (r)\,=\,0$ for $r\, \ge \,R$

Where $r$ is the distance from the centre of the charge distribution $\rho _0$ is a constant. The electric field at an internal point $(r < R)$ is

An early model for an atom considered it to have a positively charged point nucleus of charge $Ze$, surrounded by a uniform density of negative charge up to a radius $R$. The atom as a whole is neutral. For this model, what is the electric field at a distance $r$ from the nucleus?

$(a)$ Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$\left( E _{2}- E _{1}\right) \cdot \hat{ n }=\frac{\sigma}{\varepsilon_{0}}$

where $\hat{ n }$ is a unit vector normal to the surface at a point and $\sigma$ is the surface charge density at that point. (The direction of $\hat { n }$ is from side $1$ to side $2 .$ ) Hence, show that just outside a conductor, the electric field is $\sigma \hat{ n } / \varepsilon_{0}$

$(b)$ Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

A solid metal sphere of radius $R$ having charge $q$ is enclosed inside the concentric spherical shell of inner radius $a$ and outer radius $b$ as shown in figure. The approximate variation electric field $\overrightarrow{{E}}$ as a function of distance $r$ from centre $O$ is given by