MHT CET · Physics · Electrostatics
A charge 'Q' \(\mu\) C is placed at the centre of a cube. The flux through one face and two
opposite faces of the cube is respectively
- A \(\frac{\mathrm{Q}}{6 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{3 \epsilon_{0}} \mu \mathrm{Vm}\)
- B \(\frac{\mathrm{Q}}{12 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{\epsilon_{0}} \mu \mathrm{Vm}\)
- C \(\frac{\mathrm{Q}}{6 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{2 \epsilon_{0}} \mu \mathrm{Vm}\)
- D \(\frac{\mathrm{Q}}{12 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{3 \epsilon_{0}} \mu \mathrm{Vm}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{Q}}{6 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{3 \epsilon_{0}} \mu \mathrm{Vm}\)
Step-by-step Solution
Detailed explanation
By using Gauss's Law.
It is given as
\(\Phi=\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_{0}}\)
Now, the flux passing through all the six surfaces would be
\(\Phi=6 \phi=\frac{q}{\epsilon_{0}}\)
And the flux passing through each surface would be
\(\phi=\frac{q}{6 \in_{0}}\)
It is given as
\(\Phi=\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_{0}}\)
Now, the flux passing through all the six surfaces would be
\(\Phi=6 \phi=\frac{q}{\epsilon_{0}}\)
And the flux passing through each surface would be
\(\phi=\frac{q}{6 \in_{0}}\)
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