Q.

Find the flux due to the electric field through the curved surface (R is radius of curvature)

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a

a−0,b−0,c−2πR2E

b

a−πR2E,b−0,c−2πR2E

c

a−0,b−0,c−0

d

a−πR2E,b−πR2E,c−2πR2E

answer is B.

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Detailed Solution

(i) ϕ=EAcos⁡θ;ϕ=ϕCurvd+ϕtase  ϕ=EAcos⁡90∘+EA×cos⁡180∘;ϕ=E×πR2  iiϕ=0 the same amount of flux enters and comes out form the curved surfacce  (iii) ϕ=ϕCurved +ϕBase =E×Acos0∘+EAcos⁡90∘=EA;ϕ=E×2πR2;ϕ=2πR2E
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