A sphere of radius R has a charge Q. A concen | vedclass

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(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge $Q_{enclosed}$ enclosed by the surfa

Vedclass · Accepted Answer

$\frac{Q}{\epsilon_0}$

Vedclass · Answer

(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge $Q_{enclosed}$ enclosed by the surface divided by the permittivity of free space $\epsilon_0$.Mathematically,$\phi = \frac{Q_{enclosed}}{\epsilon_0}$.In this problem,the Gaussian surface is a sphere of radius $2R$,which encloses the entire charge $Q$ present on the inner sphere of radius $R$.Therefore,$Q_{enclosed} = Q$.Substituting this into Gauss's Law,we get $\phi = \frac{Q}{\epsilon_0}$.

$A$ sphere of radius $R$ has a charge $Q$. $A$ concentric spherical Gaussian surface of radius $2R$ is drawn. The electric flux associated with the Gaussian surface is ........

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