A point charge of 10 X 10^-6 C is at the centre of a cubical Gaussian surface of sides 0.5 m. What is the flux for the surface?
Answer: 1.12 X 10^5 Nm^2/C
Explanation
We know that for a Gaussian surface, flux = q/ε
Here, q = 10^-6 C and ε = 8.85 X 10^-12 C/Nm^2
Therefore, flux = 10^-6/8.85 X 10^-12
= 1.12 X 10^5 Nm^2/C.
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