A point charge causes an electric flux of

Question:

A point charge causes an electric flux of $-1.0 \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}$ to pass through a spherical Gaussian surface of $10.0 \mathrm{~cm}$ 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?

Solution:

(a) Electric flux, $\Phi=-1.0 \times 10^{3} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$

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., $-10^{3} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$.

(b) Electric flux is given by the relation,

$\phi=\frac{q}{\epsilon_{0}}$

Where,

q = Net charge enclosed by the spherical surface

$\epsilon_{0}=$ Permittivity of free space $=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$\therefore q=\phi \in_{0}$

$=-1.0 \times 10^{3} \times 8.854 \times 10^{-12}$

$=-8.854 \times 10^{-9} \mathrm{C}$

$=-8.854 \mathrm{nC}$

Therefore, the value of the point charge is $-8.854 \mathrm{nC}$.