The electric field components in a region are $E_x = \alpha x^{1/2}$, $E_y = E_z = 0$, where $\alpha

Question

The electric field components in a region are $E_x = \alpha x^{1/2}$, $E_y = E_z = 0$, where $\alpha = 800	ext{ N/(C m}^{1/2})$. A cube of side $a = 0.1	ext{ m}$ is placed such that its left face is at $x = 0.1	ext{ m}$. What is the charge enclosed by the cube?

Professor PC Thomas Classes · Answer

Net flux $\Phi = 1.05	ext{ N m}^2	ext{/C}$. By Gauss's Law, $q = \epsilon_0 \Phi = 8.854 	imes 10^{-12} 	imes 1.05 \approx 9.27 	imes 10^{-12}	ext{ C}$.