Register to reply 
Potential of Concentric Spherical Insulator and Conductor 
Share this thread: 
#1
Feb712, 12:22 AM

P: 14

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 coordinate 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)(1c)) + (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. Thanks in advance! 


#2
Feb712, 01:09 AM

HW Helper
Thanks
P: 10,365

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 


#3
Feb712, 07:14 PM

P: 14

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)? 


#4
Feb812, 12:57 AM

HW Helper
Thanks
P: 10,365

Potential of Concentric Spherical Insulator and Conductor
ehild 


#5
Feb812, 09:08 AM

P: 14

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)? 


#6
Feb812, 10:25 PM

HW Helper
Thanks
P: 10,365

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 


#7
Feb912, 12:57 PM

P: 14

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. 


#8
Feb912, 04:22 PM

HW Helper
Thanks
P: 10,365

The change of the potential from infinity to r=a is equal to the integral of the electric field between infinity and r=a.
[tex]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)[/tex] and the result of the integration is the same you wrote. ehild 


Register to reply 
Related Discussions  
Potential of Concentric Spherical Insulator and Conductor....Please Help  Introductory Physics Homework  7  
Power draw of a spherical conductor surrounded by an insulator & a conducting shell  Introductory Physics Homework  7  
Potential of Concentric Cylindrical Insulator and Conducting Shell  Advanced Physics Homework  2  
Potential of Concentric Cylindrical Insulator and Conducting Shell  Advanced Physics Homework  0  
Potential of Concentric Cylindrical Insulator and Conducting Shell.....Please Help  Introductory Physics Homework  5 