Given, φ = −6 × 103 N m2 C−1 ; r = 10 cm = 10 × 10−2 m
(i) If the radius of the Gaussian surface is doubled, the electric flux through the new surface will be the same, as it depends only on the net charge enclosed within and it is independent of the radius.
∴ φ = −6 × 103 N m2 C−1
(ii) ∴ φ = q/ε0 or q = −(8.85 × 10−12 × 6 × 103) or q = − 5.31 × 10−8 C