Question

Question 1.20: A point charge causes an electric flux of −1.0 × 10 3 Nm 2 /C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Class 12 - Physics - Electric Charges And Fields Electric Charges And Fields Page-48

Answer 1

(a) if the radius of the Gaussian surface we're doubled , then electric flux remains constant.
because electric flux depends upon charge inclosed into the Gaussian surface and permittivity of medium.
so, if we increase the radius of Gaussian surface , charge inclosed by it remains the same and hence the electric flux linked with Gaussian surface also remains the same.

(b) $\bf{\phi=\frac{q}{\epsilon_0},or,q=\phi.\epsilon_0}$
here, q = -1 × 10³ × 8.85 × 10^-12
= -8.85 × 10^-9 C
= -8.85 nC [ 1 nanoC = 10^-9C ]
hence, q = -8.85 nC

Answer 2

Answer:

Explanation:

(a) Electric flux, Φ = −1.0 × 103 N m2/C

Radius of the Gaussian surface,

r = 10.0 cm

Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −103 N m2/C.

(b) Electric flux is given by the relation,

Where,

q = Net charge enclosed by the spherical surface

∈0 = Permittivity of free space = 8.854 × 10−12 N−1C2 m−2

∴

= −1.0 × 103 × 8.854 × 10−12

= −8.854 × 10−9 C

= −8.854 nC

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