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

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<r<6 E1(r) is given by

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<r<12 E2(r) is given by

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<r<50cm is given by

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

tallinn tallinn · Answer 2 · 2022-04-12T14:04:04+02:00

When 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 Q is = 68π*(0.06³) C

What is the Coulomb's law?

The Coulomb's law for any spherically symmetric charge distribution, They are says that if the charge is within the Gaussian surface of radius r, it contributes as if it were at the center, and also if the charge is outside the surface When it does not contribute at all.

Also, This fact is considered to solve the following circumstances;

A) for 0<r<6 E1(r) is given by

Then, E1(r) = (0 →r)k∫[(4πx² ρ1dx)/(r²)] = [{(4πk ρ1)/(r²)}*(r³ /3)]

Now, = [{(4πk ρ1)/3}*r] = [4/3)*π*9*10⁹*(-5)*r] V/m

After that, E1 = -(60π*10⁹)*r V/m

And, B) for 6<r<12 E2(r) is given by

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

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

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

C) 12cm<r<50cm is given as per question

After that, E3(r) = kQ/r², where

Then, 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

Therefore, Q = 68π*(0.06³) C

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