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A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

Asked in CBSE AIPMT 2011 · Gauss's law and enclosed charge

Answer: (2) remain the same

Step-by-step solution

Gauss's law depends only on the charge enclosed:

φ=(q_enclosed)/(ε₀)=Q/(ε₀)

Doubling the radius does not change what is inside, so the flux is unchanged.

It is worth seeing the cancellation explicitly. At radius r,

E=1/(4πε₀)Q/(r²) and the area is 4π r²

φ=E×4π r²=Q/(ε₀)

The field falls to a quarter and the area grows fourfold; the product keeps no memory of r at all.

Why the other options are wrong

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