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Question

(C) According to Gauss's Law,the net electric flux $\phi$ through a closed surface is given by $\phi = \oint E \cdot dA = \frac{q_{en}}{\varepsilon_0}

Vedclass · Accepted Answer

$4\pi \varepsilon_0 A a^3$

Vedclass · Answer

(C) According to Gauss's Law,the net electric flux $\phi$ through a closed surface is given by $\phi = \oint E \cdot dA = \frac{q_{en}}{\varepsilon_0}$.Given the electric field $E = Ar$ is acting radially outward,for a spherical surface of radius $a$,the electric field at the surface is $E = Aa$.The surface area of the sphere is $S = 4\pi a^2$.Since the field is uniform over the surface of the sphere and directed radially outward,the flux is $\phi = E 	imes S = (Aa) 	imes (4\pi a^2) = 4\pi A a^3$.Equating this to Gauss's Law: $4\pi A a^3 = \frac{q}{\varepsilon_0}$.Therefore,the charge contained is $q = 4\pi \varepsilon_0 A a^3$.

The electric field in a certain region is acting radially outward and is given by $E = Ar$. $A$ charge contained in a sphere of radius $a$ centered at the origin of the field will be given by:

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