A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of - 5.0C/m3. The outer layer has a uniform charge density of + 8.0C/m3 and extends from an inner radius of 6.0 cm to an outer radius of 12.0 cm.

A) Determine the electric field for 0 B) Determine the electric field for 6.0cm C) Determine the electric field for 12.0cm

Question

Carissa Parks

Physics

17 January, 02:07

A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of - 5.0C/m3. The outer layer has a uniform charge density of + 8.0C/m3 and extends from an inner radius of 6.0 cm to an outer radius of 12.0 cm.

A) Determine the electric field for 0 B) Determine the electric field for 6.0cm C) Determine the electric field for 12.0cm

+3

Answers (1)

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Elena Mayer · Answer 1

Q = 68π * (0.06³) C

Explanation:

Coulomb's law for any spherically symmetric charge distribution, says that if charge is within the Gaussian surface of radius r, it contributes as if it were at the center, and if charge is outside the surface, it does not contribute at all. This fact is considered to solve the following conditions;

A) for 0
E1 (r) = (0 →r) k∫[ (4πx² ρ1dx) / (r²) ] = [{ (4πk ρ1) / (r²) } * (r³ / 3) ] = [{ (4πk ρ1) / 3}*r] = [4/3) * π*9*10⁹ * (-5) * r] V/m

E1 = - (60π*10⁹) * r V/m

B) for 6
E2 (r) = -[{ (4/3) π*5*9*10⁹ (0.06) ³}/r²] + (6 →r) k∫[ (4πx² ρ2dx) / (r²) ] or

E2 = - [{129560000π}/r²] + [{ (4πk ρ2) / (r²) }*{ (r³ / 3) - 0.06³/3) ] or

= - [{129560000π}/r²] + [{ (4πk*8) / (3r²) }*{ (r³-0.06³]

C) 12cm
E3 (r) = kQ/r², where

Q = (4π/3) * [-5*0.06³ + 8 * (0.12³ - 0.06³) ] = (4π/3) * [8 * (0.12³) - 13*0.06³) ] = (4π/3) * (0.06³) * [ (64-13) ] Core

Q = 68π * (0.06³) C