Electrostatic potential energy for concentric spheres

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SUMMARY

The discussion focuses on calculating the total energy stored in the electric field between two concentric metal spheres with radii r_1 = 10 cm and r_2 = 10.5 cm. The inner sphere has a charge of Q = 5 nC, while the outer sphere carries a charge of -Q. Participants agree that treating the spheres as parallel flat slabs simplifies the problem, akin to a capacitor, allowing the use of the formula U = (1/2)QV to find the energy stored. The electric field inside the inner sphere is confirmed to be zero, reinforcing the validity of this approach.

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  • Understanding of electric fields and Gauss' Law
  • Familiarity with the concept of capacitors
  • Knowledge of energy storage in electric fields
  • Basic calculus for potential difference integration
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Two concentric metal spheres have radii r_1 = 10 cm and r_2 = 10.5 cm. The inner sphere has a charge of Q = 5 nC spread uniformly on its surface, and the outer sphere has charge -Q on its surface. (a) calculate the total energy stored in the electric field inside the spheres Hint: You can treat the spheres essentially as parallel flat slabs separated by 0.5 cm why?



\phi = 4\pi kQ
U=qV/2




First of all, I don't know why treating the spheres as slabs will help, but since that's the hint, I'm looking for a way to do it. I can show with Gauss' Law that the electric field inside the inner sphere is 0, so that kind of makes them like slabs. Is that enough justification and why?
 
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HINT: Think capacitors. How do you find the energy stored in a capacitor?

Two charged slabs separated by some distance d, is essentially a capacitor.
This is why treating the spheres as flat surfaces will help. The curvature will not really affect the situation, it is essentially a capacitor, whether spherical or flat.
 
G01 said:
HINT: Think capacitors. How do you find the energy stored in a capacitor?

Two charged slabs separated by some distance d, is essentially a capacitor.
This is why treating the spheres as flat surfaces will help. The curvature will not really affect the situation, it is essentially a capacitor, whether spherical or flat.

I suppose I'll buy it just because the electric field ends up being constant like with two plates. So to find the energy I just do U=(1/2)QV. I suppose I could calculate the potential difference by integrating the electric field over that 0.5 cm distance. Would that be the way to do it?
 
I think its safe to assume that the field is constant within the capacitor. You shouldn't have to integrate, unless you want the practice of course:smile:

I would go about this using the formula for energy stored in an electric field, which is:

U = 1/2 C V^2
 
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