Question

A point charge causes an electric flux of –1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred onthe charge. (a) If the radius of the Gaussian surface were doubled,how much flux would pass through the surface? (b) What is thevalue of the point charge?

Answer 1

Given: The magnitude of electric flux is −1.0× 10 3   Nm 2 /C , the radius of Gaussian surface is 10 cm.

(a)

The net electric flux through a Gaussian surface is given as,

ϕ= q ε 0 (1)

Where, the charge placed at the center of the cube is q and the net electric flux through all six faces of the cube is ϕ T .

The magnitude of permittivity of free space is,

ε 0 =8.854× 10 −12   C 2 / Nm 2

The above equation shows that the electric flux passing through a surface depends upon the magnitude of net charge enclosed inside the surface. It does not depend on the size of the body. There is no net change on electric flux due to change in radius of Gaussian surface.

Thus, if the radius of Gaussian surface were doubled, the net electric flux passing through Gaussian spherical region remains constant as ϕ=−1.0× 10 3   Nm 2 /C .

(b)

By substituting the given values in equation (1) , we get

−1.0× 10 3 = q 8.854× 10 −12 q=−1.0× 10 3 ×8.854× 10 −12 =−8.854× 10 −9  C =−8.854 nC

Thus, the value of point charge is −8.854 nC.