Q: Let ρ(r)=Qr/(π R^{4} ) be the charge density distribution for a solid sphere of radius R and total charge Q. For a point P inside the sphere at a distance r_{1} from the centre of the sphere, the magnitude of electric field is

$\displaystyle (a)\frac{Q}{4\pi \epsilon_0 r_1^2} $

$\displaystyle (b)\frac{Q r_1^2}{4\pi \epsilon_0 R^4} $

$\displaystyle (c)\frac{Q r_1^2}{3 \pi \epsilon_0 R^4} $

$\displaystyle (d) Zero $

Ans: (c)

Sol: Consider a Gaussian Surface of radius r_{1} .

According to Gauss’s Law ,

$\displaystyle \oint \vec{E}.\vec{ds} = \frac{q_{in}}{\epsilon_0} $

$\displaystyle \oint E ds cos0 = \frac{1}{\epsilon_0}\rho(r)\times dV $

$\displaystyle E \oint ds = \frac{1}{\epsilon_0}\frac{Q r_1}{\pi R^4}\times (\frac{4}{3}\pi r_1^3) $

$\displaystyle E (4\pi r_1^2) = \frac{1}{\epsilon_0}\frac{Q r_1}{\pi R^4}\times (\frac{4}{3}\pi r_1^3) $

$\displaystyle E = \frac{Q r_1^2}{3 \pi \epsilon_0 R^4} $