Assignment-2

AIEEE Tutorials › Electricity and magnetism questions › Assignment-2

#Physics #Electricity and magnetism #Electrostatics

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Sourish Ghosh ·2014-04-21 10:33:36

E(r) = 2Ï€A(r² - a²) + Q4Ï€r²Îµ
â†’ E(r) = A2Îµ - Aa²2r²Îµ + Q4Ï€r²Îµ
For E(r) to be independent of r,
- Aa²2r²Îµ + Q4Ï€r²Îµ = 0

â†’ (C)

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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 11:32:44
Niraj kumar Jha @aditya sourish has found the net charge by integrating (4A(pi)r^2/r)dr.@sourish thanks!
Upvote·0· Reply ·2014-04-21 18:12:45
Aditya Agarwal Oh ya! Thanks. :)
Upvote·0· Reply ·2014-04-21 22:09:26
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