Let $ρ (r) = \frac{Q}{π R^{4}} r$ 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 r₁ from the centre of the sphere, the magnitude of electric field is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{Q r_{1}^{2}}{3 π ε_{0} R^{4}}$

$\frac{Q r_{1}^{2}}{4 π ε_{0} R^{4}}$

$\frac{Q}{4 π ε_{0} r_{1}^{2}}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Charge enclosed by the sphere of radius is r₁
$Q^{'} = \int_{0}^{r_{1}} ρ (4 {πr}^{2}) dr = \int_{0}^{r_{1}} \frac{Q}{{πR}^{4}} r (4 {πr}^{2}) dr \Rightarrow Q^{'} = \frac{Q}{{πR}^{4}} (4 π) {[\frac{r^{4}}{4}]}_{0}^{r_{1}} = \frac{{Qr}_{1}^{4}}{R^{4}}$
$∴ E.F.at P is E = \frac{1}{4 {πε}_{0}} \cdot \frac{Q^{'}}{r_{1}^{2}} = \frac{1}{4 {πε}_{0}} \cdot \frac{Q \frac{r_{1}^{4}}{R^{4}}}{r_{1}^{2}} \Rightarrow E = \frac{{Qr}_{1}^{2}}{4 {πε}_{0} R^{4}}$