 # A uniformly charged conducting sphere of 2.4 m `
Question:

A uniformly charged conducting sphere of $2.4 \mathrm{~m}$ diameter has a surface charge density of $80.0 \mu \mathrm{C} / \mathrm{m}^{2}$.

(a) Find the charge on the sphere.

(b) What is the total electric flux leaving the surface of the sphere?

Solution:

(a) Diameter of the sphere, d = 2.4 m

Radius of the sphere, r = 1.2 m

Surface charge density, $\sigma=80.0 \mu \mathrm{C} / \mathrm{m}^{2}=80 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2}$

Total charge on the surface of the sphere,

Q = Charge density × Surface area

$=\sigma \times 4 \pi r^{2}$

$=80 \times 10^{-6} \times 4 \times 3.14 \times(1.2)^{2}$

$=1.447 \times 10^{-3} \mathrm{C}$

Therefore, the charge on the sphere is $1.447 \times 10^{-3} \mathrm{C}$.

(b) Total electric flux $\left(\phi_{\text {Total }}\right)$ leaving out the surface of a sphere containing net charge $Q$ is given by the relation,

$\phi_{\text {TOtal }}=\frac{Q}{\epsilon_{0}}$

Where,

0 = Permittivity of free space

$=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$Q=1.447 \times 10^{-3} \mathrm{C}$

$\phi_{\text {Total }}=\frac{1.44 \times 10^{-3}}{8.854 \times 10^{-12}}$

$=1.63 \times 10^{8} \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^{2}$

Therefore, the total electric flux leaving the surface of the sphere is $1.63 \times 10^{8} \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^{2}$.