Question

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 /m2 . (a) find the charge on the sphere. (b) what is the total electric flux leaving the surface of the sphere? Ans. (a) diameter of the sphere, d = 2.4 m

Answer 1

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−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.

Answer 2

$\bigstar$ Question:

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of $\sf 80.0 \: \mu C/m^{2}$

$\bigstar$ Given:

• Diameter of the sphere = 2.4 m
• Surface charge density = $\sf 80.0 \: \mu C/m^{2}$

$\bigstar$ To Find:

• The charge on the sphere.
• The total electric flux leaving the surface of the sphere.

$\bigstar$ Solution:

Given,  The diameter of the sphere = 2.4 m

The radius of the sphere = 1.2 m

Surface charge density = $\sf 80.0 \: \mu C/m^{2} = 80 \times 10^{-6} \: C/m^{2}$

Q = Charge density × Surface area

$\implies \sf \sigma = 4\pi r^{2}$

$\implies \sf 80 \times 10^{-6} \times 4 \times 3.14 \times (1.2)^{2}$

$\implies \boxed{\sf 1.447 \times 10^{-3} \: C}$

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

Total electric flux leaving out the surface of a sphere containing net charge Q is given by the relation,

$\sf \phi \: Total = \dfrac{q}{\epsilon_0}$

$\sf \epsilon_0 = Permitivity \: of \: free \: space = 8.854 \times 10^{- 12}\: n^{- 1}\: c^{2} \: m^{ -2}$

$\sf Q= \boxed{\sf 1.63 \times 10^{8}\: N\: C^{-1} \: m^{2}}$

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