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(A) According to Gauss's Law,the electric flux $\phi$ through any closed surface enclosing a charge $q$ is given by $\phi = \oint \vec{E} \cdot d\vec{

Vedclass · Accepted Answer

$20$

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(A) According to Gauss's Law,the electric flux $\phi$ through any closed surface enclosing a charge $q$ is given by $\phi = \oint \vec{E} \cdot d\vec{S} = \frac{q}{\varepsilon_0}$.This equation shows that the electric flux through a closed surface depends only on the net charge enclosed by the surface and is independent of the shape or size of the surface.Given that the flux through the $10 \ cm$ radius sphere is $20 \ Vm$,we have $\frac{q}{\varepsilon_0} = 20 \ Vm$.Since the charge $q$ enclosed by the $20 \ cm$ radius sphere is the same as that enclosed by the $10 \ cm$ radius sphere,the flux remains unchanged.Therefore,the flux through the $20 \ cm$ radius sphere is $20 \ Vm$.

$A$ point charge is distributed in a small volume. The electric flux through a spherical surface of radius $10 \ cm$ enclosing the charge is $20 \ Vm$. What is the electric flux through a concentric spherical surface of radius $20 \ cm$?

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