Question

Let $\rho (r) = \dfrac{Q}{{\pi {R^4}}}r$ 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 center of sphere, the magnitude of electric field is
A. $0$
B. $\dfrac{Q}{{4\pi {\varepsilon _0}{r_1}^2}}$
C. $\dfrac{{Q{r_1}^2}}{{4\pi {\varepsilon _0}{R^4}}}$
D. $\dfrac{{Q{r_1}^2}}{{3\pi {\varepsilon _0}{R^4}}}$

Answer 1

Hint: In order to solve this question we need to understand gauss law for electro statistics. Gauss law states that flux through any closed surface is equal to net charge enclosed by surface divided by ${\varepsilon _0}$ also the net flux equal to net electric field crossing unit surface area and it is mathematically defined as $\phi = \oint {\vec E.d\vec S} $ and the gauss law can be derived from coulomb law for electro statistics. If the charge is distributed in volume, then volume charge density is defined as charge distributed per unit volume.

Complete step by step answer:
Given volume charge distribution, $\rho (r) = \dfrac{Q}{{\pi {R^4}}}r$.
Consider a Gaussian sphere of radius ${r_1}$ inside sphere of radius $R$.
For charge enclosed by this Gaussian sphere is given by, $q = \int {\rho dV} $.
For spherical shape the surface volume element is defined as, $dV = 4\pi {r^2}dr$.
Putting values we get,
$q = \int_0^{{r_1}} {\dfrac{{Qr}}{{\pi {R^4}}}(4\pi {r^2})dr} $
$\Rightarrow q = \dfrac{{4Q}}{{{R^4}}}\int_0^{{r_1}} {{r^3}dr} $
since, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
$q = \dfrac{{4Q}}{{{R^4}}}(\dfrac{{{r_1}^4}}{4})$
$\Rightarrow q = \dfrac{{Q{r_1}^4}}{{{R^4}}}$

Therefore from gauss law net flux enclosed by surface is given by,
$\phi = \dfrac{q}{{{\varepsilon _0}}}$
Putting values we get,
$\phi = \dfrac{{Q{r_1}^4}}{{{R^4}{\varepsilon _0}}} \to (i)$
Also from definition of flux,
$\phi = \oint {\vec E.d\vec S} $
Also $E$ and $dS$ is in the same direction along $\hat r$. Since $E$ is uniform so,
$\phi = E\oint {dS} $
$\Rightarrow \phi = E(4\pi {r_1}^2) \to (ii)$
Equating equation $(i)\& (ii)$ we get,
$E(4\pi {r^2}) = \dfrac{{Q{r_1}^4}}{{{R^4}{\varepsilon _0}}}$
$\Rightarrow E = \dfrac{{Q{r_1}^2}}{{4\pi {\varepsilon _0}{R^4}}}$
$\therefore \vec E = \dfrac{{Q{r_1}^2}}{{4\pi {\varepsilon _0}{R^4}}}\hat r$

So the correct option is C.

Note: It should be remembered that Gauss law is only applicable for closed surfaces.There are two types of surface one is open surface and other is closed surface, open surface is defined as surface not enclosing any volume but a closed surface is one which encloses. For example, a cuboid is a closed surface while a rectangle is an open surface. Also gauss law is defined for uniform electric fields.