Insulating charged sphere in conducting shell and electric field

  • #1
2
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1. A solid sphere of radius a = 12.8 cm is concentric with a spherical conducting shell of inner radius b = 37.1 cm and outer radius c = 39.1 cm. The sphere has a net uniform charge q1 = 9.00×10-6 C. The shell has a net charge q2 = -q1. Find expressions for the electric field, as a function of the radius r, between the sphere and the shell (a < r < b). Evaluate for r = 25.0 cm.


2. ∫E.dA = Q/ε0
ρ = Q/Volume



3. OK, so I have the charge density of the insulting sphere, which I'm calling ρ. My Gaussian surface is a sphere, so it's area in this case would be 4∏(0.25m)^2
I know how to find the electric field inside of the insulating sphere, but not between the insulating sphere and the conducting shell, which is what this problem is asking for. I tried ignoring the conducting shell and just using the equation:
E = (ρa^2)/(3ε0r^2)
It's telling me that the answer is wrong. What am I doing wrong? I assume it has something to do with the charge on the inner surface of the conducting shell, but I don't know what to do with that.
 

Answers and Replies

  • #2
1. A solid sphere of radius a = 12.8 cm is concentric with a spherical conducting shell of inner radius b = 37.1 cm and outer radius c = 39.1 cm. The sphere has a net uniform charge q1 = 9.00×10-6 C. The shell has a net charge q2 = -q1. Find expressions for the electric field, as a function of the radius r, between the sphere and the shell (a < r < b). Evaluate for r = 25.0 cm.


2. ∫E.dA = Q/ε0
ρ = Q/Volume



3. OK, so I have the charge density of the insulting sphere, which I'm calling ρ. My Gaussian surface is a sphere, so it's area in this case would be 4∏(0.25m)^2
I know how to find the electric field inside of the insulating sphere, but not between the insulating sphere and the conducting shell, which is what this problem is asking for. I tried ignoring the conducting shell and just using the equation:
E = (ρa^2)/(3ε0r^2)
It's telling me that the answer is wrong. What am I doing wrong? I assume it has something to do with the charge on the inner surface of the conducting shell, but I don't know what to do with that.


Hi Kyle,welcome to PF.

The equation in red is wrong. Why do you use the charge density instead of Gauss Law with the enclosed charge q1=9.00×10-6 C?

ehild
 
  • #3
Thanks! I guess I was overthinking. Instead of using the whole Q = (4/3)πr^2*ρ, I just had to use the charge given to me.
 

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