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Asked by Annu Last Modified
Answer 
Apoorva Purohit
(a) Diameter of the sphere, d = 2.4 m
Radius of the sphere, r = 1.2 m
Surface charge density, 
= 80.0 μC/m2 = 80 × 10−6 C/m2
Total charge on the surface of the sphere,
Q = Charge density × Surface area
=
= 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 (
) leaving out the surface of a sphere containing net charge Q is given by the relation,
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
Q = 1.447 × 10−3 C
= 1.63 × 108 N C−1 m2
Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 108 N C−1 m2.
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