Question 22

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?

Answer

(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

∈_{0} = 8.854 × 10^{−12} N^{−1}C^{2} m^{−2}

Q = 1.447 × 10^{−3} C

= 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}.

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Steven
2019-11-24 07:32:59

Thank you it was really useful

- NCERT Chapter

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