MHT CET · Physics · Electrostatics
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
- A \(\frac{Q}{6 \epsilon_0}, \frac{Q}{3 \epsilon_0}\)
- B \(\frac{Q}{3 \epsilon_0}, \frac{Q}{2 \epsilon_0}\)
- C \(\frac{Q}{12 \epsilon_0}, \frac{Q}{6 \epsilon_0}\)
- D \(\frac{Q}{\epsilon_0}, \frac{Q}{2 \epsilon_0}\)
Answer & Solution
Correct Answer
(A) \(\frac{Q}{6 \epsilon_0}, \frac{Q}{3 \epsilon_0}\)
Step-by-step Solution
Detailed explanation
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.
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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