A spherical ball of radius R is uniformly charged. The surrounding space of ball is charged with volume charge density $\frac{α}{r}$ , where $α$ is a constant and $r$ is the distance from the centre of the ball. The value of charge q on ball for which the strength of electric field outside the ball is independent of $r$ . (The relative permittivites of the ball and the surrounding space are assumed to be equal to unity)

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

Electric field out-side the ball is $E = \frac{2 α}{ε_{0}}$

Charge on ball is $q = 2 π α R^{2}$

Electric field out-side the ball is $E = \frac{α}{2 ε_{0}}$

Charge on ball is $q = π α R^{2}$

answer is B, D.

(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

Question Image

$\int d q = \int \frac{α}{r} 4 π r^{2} d r q = 4 π α \cdot {[\frac{r^{2}}{2}]}^{r} = 2 π α [r^{2} - R^{2}] \frac{k (Q + q)}{r^{2}} = E E = \frac{k (Q + 2 π α (r^{2} - R^{2}))}{r^{2}} E = \frac{k Q}{r^{2}} + 2 π α k - \frac{2 π α k R^{2}}{r^{2}}$