A uniformly charged conducting sphere of $2.4\; m$ diameter has a surface charge density of $80.0\; \mu \,C/m^2$.

$(a)$ 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, $\sigma=80.0\, \mu\, C / m ^{2}=80 \times 10^{-6} \,C / m ^{2}$

Total charge on the surface of the sphere, $Q=$ Charge density $\times$ 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} \,C$

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

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

$\phi_{\text {Total }}=\frac{Q}{\varepsilon_{0}}$

Where, $\varepsilon_{0}=$ Permittivity of free space $=8.854 \times 10^{-12} \,N ^{-1} \,C ^{2} \,m ^{-2}$

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

$\therefore \phi_{\text {Total }}=\frac{1.447 \times 10^{-3}}{8.854 \times 10^{-12}}$

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

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