Question

An isolated charged conducting sphere has a radius R = 14.0 cm. At a distance of r = 24.0 cm from the center of the sphere the electric field due to the sphere has a magnitude of E = 4.90 ✕ 104 N/C. (a) What is its surface charge density (in µC/m2)? µC/m2 (b) What is its capacitance (in pF)? pF (c) What If? A larger sphere of radius 30.0 cm is now added so as to be concentric with the first sphere. What is the capacitance (in pF) of the two-sphere system? pF

Answer 1

here isolated conducting sphere is given,

we have formula of conductiong sphere for finding electric field outside the sphere, when r > R

E = Q / 4πε₀ r² ( when r > R )

here in example, r = 24.0 cm = 0.24 m and R = 14 cm = 0.14 m. so here r>R

here E = 4.90 ✕ 10⁴ N/C , ε₀ =8.85×10⁻¹² F⋅m⁻¹ , r= .24 m

4.90 ✕ 10⁴ = Q/ 4 ✕ 3.1428✕ 8.85×10⁻¹² ✕ ( 0.24)²

4.90 ✕ 10⁴ = Q / 6.40829× 10⁻¹²

4.90 ✕ 10⁴ ✕ 6.40829× 10⁻¹² = Q

Q = 31.600421  ✕ 10^-8 C

Q = 3.16 ✕ 10^-7 C

(a)

here we want to find the surface charge density,

surface charge density = charge / area

This charge is uniformly distributed over the surface of the sphere of radius R thus the charge per unit area of the sphere

here area of sphere = 4 πR²

  = 4 * 3.1428 *(0.14)²
=0.246395 m ²

surface charge density, ρs = charge / area

= 3.16 ✕ 10^-7 /  0.246395

  ρs = 12.8249  ✕ 10^-7 C/ m²

but here we want to find ans in µC/m² , so 1 µC/m² = 10^-6 C/m²

so here we have to convert answer in the form of 10^-6,

    ρs = 12.8249  ✕ 10^-7 C/ m²

    ρs = 1.28249✕ 10^-6 C/ m²

  ρs = 1.282 ✕ 10^-6 C/ m²

ρs = 1.282  µC/m²

(b) we want to find capacitance for the sphere, we have general formula for finding capacitance of the sphere , C = 4πε₀R

here we have all value, put all value in equation.

C = 4πε₀R

= 4 ✕ 3.1428 ✕ 8.85×10⁻¹²  ✕ 0.14

C= 15.5757  ✕ 10⁻¹² F

but here we want to find answer in pF unit, so 1 pF = 10^-11 F

C = 15.5757  ✕ 10⁻¹² F , so we have to convert in the form of 10^-11

c = 1.55757  ✕ 10⁻¹¹ F

C = 1.557 pF ( 10⁻¹¹ F = 1 pF)

(C) here another sphere is added with radius r = 30 cm = 0.30 m

we have general formula for finding capacitance two sphere system,

C= 4πε₀ r₁r₂ / r₂-r₁

here r₁ = inner sphere radius  

r₂ = outer sphere radius

here r₁ = 0.14 m ( inner sphere radius )

r₂ = 0.30 m ( added outer sphere radius)

C = 4πε₀ r₁r₂ / r₂-r₁

=(4✕3.1428✕8.85✕10^-12 ✕0.14✕0.30) / (0.30-0.14)

= 4.6727 ✕10^-12 / 0.16

C = 29.20 ✕10^-12 F

but here we want to find answer in pF unit, so 1 pF = 10^-11 F

C = 29.20 ✕10^-12 F , so we have to convert in the form of 10^-11

C = 2.92✕10^-11 F

C = 2.92 pF ( 10⁻¹¹ F = 1 pF)