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=EAcosθ;ϕ=ϕCurvd+ϕtase ϕ=EAcos90∘+EA×cos180∘;ϕ=E×πR2 iiϕ=0 the same amount of flux enters and comes out form the curved surfacce (iii) ϕ=ϕCurved +ϕBase =E×Acos0∘+EAcos90∘=EA;ϕ=E×2πR2;ϕ=2πR2E
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