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?

2.45 X 10^5 Nm^2/C

1.12 X 10^5 Nm^2/C

3.12 X 10^5 Nm^2/C

None of these

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.