E(r) = 2πA(r2 - a2) + Q4πr2ε
→ E(r) = A2ε - Aa22r2ε + Q4πr2ε
For E(r) to be independent of r,
- Aa22r2ε + Q4πr2ε = 0
→ (C)
Aditya Agarwal Why have you considered it to be 2pi? Its a spherical charge distribution. So shouldnt it be 4(pi)(r^2-a^2)?
Upvote·0· Reply ·2014-04-21 11:32:44Niraj kumar Jha @aditya sourish has found the net charge by integrating (4A(pi)r^2/r)dr.@sourish thanks!
Aditya Agarwal Oh ya! Thanks. :)