Question

- An electric flux of - 6x 103 Nm²/C passes normally through a spherical
Gaussian surface of radius 10 cm due to a point charge placed at the centre.
i) what is the charge enclosed by the Gaussian surface? ii) If the radius of
the Gaussian surface is doubled, how much flux would pass through the
surface
пасе?​

Answer 1

Answer:

Gauss law states that net flux through a closed surface is 1/ permittivity times the net charge enclosed within the surface.

You know the value of flux -6 x 103, you know the value of permittivity of free space = 8.85 x 10-12

So charge enclosed = flux x permittivity = -6 x 103 x 8.85 x 10-12 = -53.1 x 10-9 = -5.31 x 10-8 C

---------

If the radius of Gaussian surface is doubled. it would have no change on the overall flux. It will remain -6 x 103 Nm2 C-1

Flux through closed surface doesn't depend on the geometry of gaussian surface. It only depends on net charge enclosed and permittivity of medium.

Answer 2

Answer:

mark me as brainliest and support me to give more valuble answers

Explanation:

When dipole moment vector is parallel to electric field vector.