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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 $Q_{	ext{enclosed}}$ enclosed by th

Vedclass · Accepted Answer

$\frac{Q}{\varepsilon_0}$

Vedclass · Answer

(A) According to Gauss's law,the total electric flux $\phi$ through any closed surface is equal to the net charge $Q_{	ext{enclosed}}$ enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.Mathematically,$\phi = \frac{Q_{	ext{enclosed}}}{\varepsilon_0}$.In this problem,the imaginary sphere of radius $2R$ encloses the sphere of radius $R$ which carries a charge $Q$.Therefore,the net charge enclosed by the imaginary sphere is $Q_{	ext{enclosed}} = Q$.Substituting this into Gauss's law,we get $\phi = \frac{Q}{\varepsilon_0}$.

$A$ sphere of radius $R$ and charge $Q$ is placed inside a concentric imaginary sphere of radius $2R$. The flux associated with the imaginary sphere is

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