Question

An electric flux of – 6 x 103 Nm2/C passes normally through a spherical Gaussian surface of a radius of 10 cm, due to a point of 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 will pass through the surface?

Answer 1

(i) φ = – 6 x 103Nm2C-1
By Gauss’s theorem, φ = \(\frac{\mathrm{q}}{\varepsilon_{0}}\)
q= \(\varepsilon_{0}\) x φ = 8.85 x 1o-12 x – 6 x 103 = – 5.31 x 10-8C

(ii) Same (- 6 x 103 Nm2C-1)