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Question attached.. [ALTS-3 2013 Q37]

Question attached..
 
[ALTS-3 2013 Q37]

Question Image
Grade:12th pass

1 Answers

Atal Tiwari
37 Points
6 years ago
By symmetry net electic field along X-axis at the centre O is zero and electric field along Y-axis will be added up.
dEY=1/4pi-epsilon-naught x dq/r2sin@
but      dq= ∂(rd@)=q/pi-r x (rd@) = q/pi d@
therefore,    EY=2 x 1/4-pi-epsilon-naught ∫pi/2(q/pi sin@ d@)/r2
                        =q/2-pi2-epsilon naught-r2.
Hence, the correct part is (a).

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