Let $P\left( r \right) = \frac{Q}{{\pi {R^4}}}r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point $P$ inside the sphere at distance $r_1$ from the centre of the sphere, the magnitude of electric field is
zero
$\frac{Q}{{4\pi {\varepsilon _0}r_1^2}}$
$\frac{{Qr_1^2}}{{4\pi {\varepsilon _0}{R^4}}}$
$\frac{{Qr_1^2}}{{3\pi {\varepsilon _0}{R^4}}}$
Obtain Coulomb’s law from Gauss’s law.
The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero ?
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 $2 \ R$ from the axis of the cylinder, is given by the expression $\frac{23 \rho R }{16 k \varepsilon_0}$. The value of $k$ is
Two concentric conducting thin spherical shells of radii $a$ and $b\ (b > a)$ are given charges $Q$ and $ -2Q$ respectively. The electric field along a line passing through centre as a function of distance $(r)$ from centre is given by
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0\times 10^{-22}\; C/m^2$. What is $E$:
$(a)$ in the outer region of the first plate,
$(b)$ in the outer region of the second plate, and
$(c)$ between the plates?