# Electrostatic potential energy for concentric spheres

• wombat7373
Where C is the capacitance and V is the potential difference. Since we can treat the spheres as parallel slabs, we can use the formula for capacitance of parallel plates:C = (epsilon_0 A)/dWhere epsilon_0 is the permittivity of free space, A is the area of the plates, and d is the distance between them. In this case, A can be calculated using the formula for the surface area of a sphere:A = 4 pi r^2Plugging in the given values, we get:C = (8.85x10^-12 * 4 pi (10x10^-2)^2)/(0.5x10^-2) = 7.08

#### wombat7373

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 teh 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

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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