The electric field in a region is radially ou | vedclass

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(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{enclosed}}{\varepsilon_0}$.Here,the

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$4\pi \varepsilon_0 A r_0^3$

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(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{enclosed}}{\varepsilon_0}$.Here,the electric field $E = A r_0$ is radially outward and uniform over the surface of a sphere of radius $r_0$.The surface area of the sphere is $S = 4\pi r_0^2$.The total electric flux $\phi$ through the sphere is $\phi = E 	imes S = (A r_0) 	imes (4\pi r_0^2) = 4\pi A r_0^3$.Equating the flux to Gauss's Law: $\frac{Q}{\varepsilon_0} = 4\pi A r_0^3$.Therefore,the enclosed charge is $Q = 4\pi \varepsilon_0 A r_0^3$.

The electric field in a region is radially outward with magnitude $E = A r_0$. The charge contained in a sphere of radius $r_0$ centered at the origin is

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