Work, energy stored in solid sphere

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SUMMARY

The discussion focuses on calculating the energy stored in a uniformly charged solid sphere with charge q and radius R. The solution involves integrating the electric field equations, specifically using the formula for electric field inside and outside the sphere. The final expression for the stored energy is derived as U = (q^2 / (4πε₀R))(3/5). The conversation also touches on the importance of distinguishing between a uniformly charged sphere and a spherical shell, as this affects the integral used in the calculations.

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Homework Statement


Find the energy stored in a uniformly charged sphere of charge q, radius R

Homework Equations





The Attempt at a Solution


Ein=\frac{qr}{4\pi\epsilon o R^3}, Eout=\frac{q}{4\pi\epsilon o r^2}... W=\int_{0}^ {R}\int_{0}^{2\pi}\int_{0}^{\pi}[\frac{qr}{4\pi\epsilon oR^3}] ^2sin\theta d\theta d\phi r^2\ dr+ \int_{R}^ {\infty }\int_{0}^{2\pi}\int_{0}^{\pi}[\frac{q}{4\pi\epsilon or^2}] ^2sin\theta d\theta d\phi r^2\ dr= 2\pi\epsilon o(\frac{q}{4\pi\epsilon o})^2(\frac{1}{5R}+\frac{1}{R})=\frac{q^2}{4\pi\epsilon o R}\frac{3}{5}

by the way the work in this case is also like the effort needed to bring the whole solid sphere in from infinity by point charges or also the stored energy?
 
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Is this a uniforms sphere of charge (i.e. as you may find in an insulator), or a spherical shell of charge (i.e. the charges are being placed on a conductor)? It affects the integral.

But yep - the electrostatic energy stored in a system of charges is the work needed to assemble them from infinity.

Nice LaTeX ... you can make a newline with a \\ to avoid running off the end of the page;
you can make subscripts with _{} like this: ##\epsilon_0## and ##E_{out}##.
trig functions are written \sin \cos etc.
 
thanks! this is a solid sphere of charge. I'm curious how you develop the mathematics for this, is it based on a physical intuition or a mathematical result of the electric field equation
 
Last edited:
OK... you seem to be using:
$$U=\int_V E^2d\tau + \int_S VEda$$
... it's a good idea to explain your process.

You ended up with: $$U=\frac{1}{2k}k^2q^2\left( \frac{1}{5R}+\frac{1}{R}\right)$$ ... where ##k=1/4\pi\epsilon_0##

Your next step is to simplify the expression.
Did you have any other questions?
 

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