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