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