A charge Q C is placed at the center of a cu | vedclass

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(A) According to Gauss's Law,the total electric flux $\phi_{total}$ through a closed surface enclosing a charge $Q$ is given by $\phi_{total} = \frac{

Vedclass · Accepted Answer

$\frac{Q}{6 \varepsilon_0}, \frac{Q}{3 \varepsilon_0}$

Vedclass · Answer

(A) According to Gauss's Law,the total electric flux $\phi_{total}$ through a closed surface enclosing a charge $Q$ is given by $\phi_{total} = \frac{Q}{\varepsilon_0}$.Since the charge is placed at the center of the cube,the flux is distributed equally among all $6$ faces due to symmetry.Therefore,the flux through one face is $\phi_{one} = \frac{\phi_{total}}{6} = \frac{Q}{6 \varepsilon_0}$.The flux through two opposite faces is the sum of the flux through each of these two faces,which is $\phi_{two} = 2 	imes \phi_{one} = 2 	imes \frac{Q}{6 \varepsilon_0} = \frac{Q}{3 \varepsilon_0}$.

$A$ charge $Q \ C$ is placed at the center of a cube. If $\varepsilon_0$ is the permittivity of vacuum,then the flux through one face and two opposite faces of the cube is respectively:

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