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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' centred at the origin of the field, will be given by
A
$4\pi\varepsilon_0Aa^2$
B
$A\varepsilon_0a^2$
C
$4\pi\varepsilon_0Aa^3$
D
$\varepsilon_0Aa^3$
Detailed Solution
The field is radial, so $\phi = \oint \vec E\cdot d\vec s = E(4\pi a^2)$.
At r = a: $\phi = Aa\times 4\pi a^2 = 4\pi Aa^3$
Gauss's law: $\phi = \frac{q}{\varepsilon_0}$
$q = 4\pi\varepsilon_0Aa^3$
At r = a: $\phi = Aa\times 4\pi a^2 = 4\pi Aa^3$
Gauss's law: $\phi = \frac{q}{\varepsilon_0}$
$q = 4\pi\varepsilon_0Aa^3$