A charge Q is enclosed by a Gaussian spherica | vedclass

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(A) According to Gauss's Law,the net electric flux $\phi$ through any closed surface is given by $\phi = \frac{Q_{	ext{enclosed}}}{\varepsilon_0}$.He

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remain the same

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(A) According to Gauss's Law,the net electric flux $\phi$ through any closed surface is given by $\phi = \frac{Q_{	ext{enclosed}}}{\varepsilon_0}$.Here,$Q_{	ext{enclosed}} = Q$. Therefore,$\phi = \frac{Q}{\varepsilon_0}$.Since the electric flux depends only on the charge enclosed by the surface and not on the shape or size of the Gaussian surface,changing the radius from $R$ to $2R$ does not affect the net electric flux.Thus,the outward electric flux will remain the same.

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

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