Consider a sphere of radius $R$ with charge density distributed as :
$\rho(r) =k r$, $r \leq R $
$=0$ for $r> R$.
$(a)$ Find the electric field at all points $r$.
$(b)$ Suppose the total charge on the sphere is $2e$ where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.
$(a)$ Let us consider a sphere $\mathrm{S}$ of radius $\mathrm{R}$ and two hypothetic spheres of radius $r<\mathrm{R}$ and $r>\mathrm{R}$. Electric field intensity for point $r<\mathrm{R}$,
$\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{S}}=\frac{1}{\epsilon_{0}} \int \rho d \mathrm{~V}$
But, volume $\mathrm{V}=\frac{4}{3} \pi r^{3}$
$d \mathrm{~V} =\frac{4}{3} \pi \times 3 r^{2} d r$
$d \mathrm{~V} =4 \pi r^{2} d r$
$\therefore \oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{S}}=\frac{1}{\epsilon_{0}} 4 \pi k \int_{0}^{r} r^{3} d r \quad[\rho=k r]$ $\therefore(\mathrm{E}) 4 \pi r^{2}=\frac{4 \pi k}{\epsilon_{0}} \cdot \frac{r^{4}}{4}$ $\therefore \mathrm{E}=\frac{1}{4 \epsilon_{0}} \cdot k r^{2}$ As charge density is positive it means the direction of $\overrightarrow{\mathrm{E}}$ is radially outwards.
An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho $. It has a spherical cavity of radius $R/2$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$, which is at a distance $2R$ from the axis of the cylinder, is given by the expression $\frac{{23\rho R}}{{16K{\varepsilon _0}}}$ .The value of $K$ is
Shown in the figure are two point charges $+Q$ and $-Q$ inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If $\sigma _1$ is the surface charge on the inner surface and $Q_1$ net charge on it and $\sigma _2$ the surface charge on the outer surface and $Q_2$ net charge on it then
Two infinitely long parallel conducting plates having surface charge densities $ + \sigma $ and $ - \sigma $ respectively, are separated by a small distance. The medium between the plates is vacuum. If ${\varepsilon _0}$ is the dielectric permittivity of vacuum, then the electric field in the region between the plates is
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.
A solid ball of radius $R$ has a charge density $\rho $ given by $\rho = {\rho _0}\left( {1 - \frac{r}{R}} \right)$ for $0 \leq r \leq R$. The electric field outside the ball is