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Q.

A cylindrical Gaussian surface of radius R and length 2R is placed in a uniform electric field of strength E→, with its two ends parallel to the field. Then outward electric flux through the Gaussian surface is

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a

2πR2E

b

zero

c

4R2E

d

2R2E

answer is C.

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Detailed Solution

Projected area, A=2R×2R=4R2∴ Outward flux =EA=4R2E
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