The electric field at a distance $r$ from the centre in the space between two concentric metallic spherical shells of radii $r_1$ and $r_2$ carrying charge $Q_1$ and $Q_2$ is $(r_1 < r < r_2)$

An infinitely long positively charged straight thread has a linear charge density $\lambda \mathrm{Cm}^{-1}$. An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of electron as a function of radius of circular path from the wire is :

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.

Let $E_1(r), E_2(r)$ and $E_3(r)$ be the respective electric fields at a distance $r$ from a point charge $Q$, an infinitely long wire with constant linear charge density $\lambda$, and an infinite plane with uniform surface charge density $\sigma$. if $E_1\left(r_0\right)=E_2\left(r_0\right)=E_3\left(r_0\right)$ at a given distance $r_0$, then

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\sigma / 2 \varepsilon_{0}\right) \hat{ n },$ where $\hat{ n }$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.