Rankers Physics

Gauss's Law: Practice Problem & Solution

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: (2015)
$A\epsilon_0 a^2$
$4\pi\epsilon_0 A a^3$
$\epsilon_0 A a^3$
$4\pi\epsilon_0 A a^2$

Solution Explained:

To solve this problem, we apply the core principles of Gauss's Law. Understanding the underlying formula is key to arriving at the correct answer below:

According to Gauss's law, the electric flux $\Phi = \oint E \cdot dA = \frac{q_{in}}{\epsilon_0}$. The electric field at radius $a$ is $E = Aa$. Therefore, flux $\Phi = (Aa)(4\pi a^2) = 4\pi A a^3$. Equating this to $\frac{q_{in}}{\epsilon_0}$, we get $q_{in} = 4\pi\epsilon_0 A a^3$.

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