Question

a particle charge of 1.7 μc is at the center of a gaussian cube 82 cm on edge. what is the net electric flux through the surface? attempts: 0 of 3 used save for later etextbook and media submit answer current attempt in progress view policies

Explanation:

Step1: Recall Gauss's law

Gauss's law states that $\varPhi_E=\frac{q_{enclosed}}{\epsilon_0}$, where $\varPhi_E$ is the electric - flux, $q_{enclosed}$ is the charge enclosed by the Gaussian surface, and $\epsilon_0 = 8.85\times10^{- 12}\ C^2/(N\cdot m^2)$ is the permittivity of free space.

Step2: Identify the enclosed charge

The charge $q_{enclosed}=1.7\ \mu C=1.7\times10^{-6}\ C$.

Step3: Calculate the electric flux

Substitute the values into Gauss's law: $\varPhi_E=\frac{1.7\times10^{-6}\ C}{8.85\times10^{-12}\ C^2/(N\cdot m^2)}\approx1.92\times10^{5}\ N\cdot m^2/C$.

Answer:

$1.92\times10^{5}$; $N\cdot m^2/C$