A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/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 μC/m^{2} = 80 × 10^{−6} C/m^{2}

Total charge on the surface of the sphere,

Q = Charge density × Surface area

= `sigma xx 4pir^2`

= 80 × 10^{−6} × 4 × 3.14 × (1.2)^{2}

= 1.447 × 10^{−3} C

Therefore, the charge on the sphere is 1.447 × 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_"total" = "Q"/in_0`

Where,

∈_{0} = Permittivity of free space

= 8.854 × 10^{−12} N^{−1 }C^{2 }m^{−2}

Q = 1.447 × 10^{−3} C

`phi_"total" = (1.44 xx 10^-3)/(8.854 xx 10^-12)`

= 1.63 × 10^{8} N C^{−1} m^{2}

Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 10^{8} N C^{−1} m^{2}.