A charge $Q \mathrm{C}$ is placed at the center of a cube. If $\varepsilon_0$ is the permittivity of vacuum…
A charge $Q \mathrm{C}$ 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
$\frac{Q}{6 \epsilon_0}, \frac{Q}{3 \epsilon_0}$
$\frac{Q}{3 \epsilon_0}, \frac{Q}{2 \epsilon_0}$
$\frac{Q}{12 \epsilon_0}, \frac{Q}{6 \epsilon_0}$
$\frac{Q}{\epsilon_0}, \frac{Q}{2 \epsilon_0}$
Solution
A charge $Q \mathrm{C}$ is placed at the center of a cube.
Total flux radiated $=\frac{Q}{\varepsilon_0}$
$\therefore$ from one face would be $\frac{Q}{6 \varepsilon_0}$ due to the six-fold symmetry of the cube.
And from two opposite faces, it would be $\frac{Q}{3 \varepsilon_0}$ due to three-fold symmetry.
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