## Potential of Concentric Spherical Insulator and Conductor

1. The problem statement, all variables and given/known data
A solid insulating sphere of radius a = 5.6 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density ρ = -159 μC/m3. Concentric with the sphere is an uncharged spherical conducting shell of inner radius b = 10.7 cm, and outer radius c = 12.7 cm.

A charge Q = 0.0724μC is now added to the conducting shell. What is V(a), the electric potential at the outer surface of the insulating sphere, now? Define the potential to be zero at infinity.

2. Relevant equations
E = kQ/(r^2)
Q = ρV

3. The attempt at a solution
I know that if the 0.0724μC was not present, the electric potential at the outer surface of the insulating sphere would be:
-kQ((1/b)-(1/c)) + (kQ/a) (*equation 1)
Using Q = ρV = -1.1696x10^-7 C.

But now that the spherical conducting shell has charged, I'm confused. Letting
0.0724μC = Q
-0.11696μC = q
I've tried:
((Q+q)/(4πεo))*(1/c) + (q/(4πεo))*((1/b)-(1-c)) + (q/(4πεo))*(1/a)
but this is wrong and I'm not sure why. I've also tried taking my answer from *equation 1 and adding that to (Q/4πεo)*(1/c) or (Q/4πεo)*(1/b) and it is still wrong.

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 Recognitions: Homework Help Use Gauss' Law to find the electric fields. Use that the electric field is zero inside the metal shell. In the picture, the charge of the insulating sphere is Q1, and the charge of the shell is distributed between the inner and outer surfaces, so as Q2+Q3=Q=0.0724μC. ehild Attached Thumbnails
 So if I make the Gaussian surface b < r < c, the Qenclosed equals zero, which means that the inner surface of the conducting shell will be equal in magnitude but opposite in direction of Q1? Q3 would be Q1+0.0724μC? So would I sum up the charges and their distances like: kQ3((1/c)-(1/a)) + kQ2((1/b)-(1/a)) + kQ1(1/a)?

Recognitions:
Homework Help

## Potential of Concentric Spherical Insulator and Conductor

 Quote by diethaltao So if I make the Gaussian surface b < r < c, the Qenclosed equals zero, which means that the inner surface of the conducting shell will be equal in magnitude but opposite in direction of Q1? Q3 would be Q1+0.0724μC? So would I sum up the charges and their distances like: kQ3((1/c)-(1/a)) + kQ2((1/b)-(1/a)) + kQ1(1/a)?
You wrote an expression and did not say what it means. How do you calculate the potential in a region r>R?

ehild

 Sorry, that expression was supposed to be what I thought would find the potential for the outer surface of the insulating sphere. Potential outside the sphere would just be the Qenclosed divided by 4πεo times the integral dr/r, right? (0.00724μC + ρV)/(4πεo) * ln(r)?

Recognitions:
Homework Help
 Quote by diethaltao Sorry, that expression was supposed to be what I thought would find the potential for the outer surface of the insulating sphere.
Explain how did you get it. It is not quite correct. In case of a spherical symmetric charge distribution the electric field outside the sphere is the same as if all the charge concentrated in the centre.

 Quote by diethaltao Potential outside the sphere would just be the Qenclosed divided by 4πεo times the integral dr/r, right? (0.00724μC + ρV)/(4πεo) * ln(r)?
No, that is not right. How are the potential and electric field related? Haven't you missed a negative sign?
What is the electric field around a metal sphere with enclosed charge Q? Is it E=kQ(enclosed)/r really? Would it satisfy Gauss' Law?

ehild

 Ah, whoops! To find potential, you integrate electric field. E = (Qouter + Qinner)/(4πεo) ∫dr/(r^2) from infinity to c, and add that to E = (Qinner)/(4πεo) ∫dr/(r^2) from b to a? So potential would be: (Qouter + Qinner)/(4πεo) * (1/c) + (Qinner)/(4πεo) * ((1/a)-(1/b)) But I don't understand where the negative sign comes into place.
 Recognitions: Homework Help The change of the potential from infinity to r=a is equal to the integral of the electric field between infinity and r=a. $$V(a)-V(\infty)=1/(4\pi \epsilon_0)\int_{\infty}^a{-Edr}=1/(4\pi \epsilon_0)\left(\int_{\infty}^c{-\frac{Q_{outer}+Q_{inner}}{r^2}dr}+\int_c^b{0 dr}+\int_b^a{-\frac{Q_{inner}}{r^2}dr}\right)$$ and the result of the integration is the same you wrote. ehild