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Question

Three charges of q1=1×106 C, q2=2×106 C and q3=3×106 C have been placed as shown. Then, the net electric flux will be maximum for the surface

A
S2
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B
S1
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C
Same for all three
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D
S3
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Solution

The correct option is B S1
According to the Gauss's law the flux of the electric field through closed surface is
ϕ=qenε0
Where, qen is the charge enclosed by the closed surface.

So, the flux will be maximum for that surface, which incloses maximum charge.

Charge enclosed by surface S1,
QS1=q1+q2
QS1=1×106+2×106
QS1=3×106 C

Charge enclosed by surface S2,
QS2=q2+q3
QS2=2×1063×106
QS2=1×106 C

Charge enclosed by surface S3,
QS3=q1+q2+q3
QS3=1×106+2×1063×106
QS3=0

So, the charge enclosed is maximum across S1
So, flux will be maximum across S1.

Hence, option (a) is correct asnwer.

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