Electrostatics, Energy of a uniformly charged sphere

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Homework Help Overview

The discussion revolves around calculating the energy stored in a uniformly charged solid sphere of radius R and charge q, using different equations from Griffiths' textbook on electrodynamics. The original poster attempts to solve the problem using three specified equations but encounters discrepancies in their results.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster describes their approach to part (a) using the potential and charge density, while parts (b) and (c) yield different results. Participants question the continuity of the potential across the sphere's surface and suggest a possible sign error in the calculations.

Discussion Status

Some participants have offered guidance regarding potential errors in the expressions for the electric potential. The original poster acknowledges a mistake in their calculations and indicates that correcting this has led to consistent results with the other parts of the problem.

Contextual Notes

The original poster mentions specific equations from the textbook and expresses uncertainty about their calculations, indicating a need for clarification on the potential and energy relationships in the context of electrostatics.

NeoDevin
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From Griffiths, Third edition Intro Electrodynamics

I've been working on this problem, and I get 2 different answers. I get the same answer for parts b and c, but not for a. I know I'm probably just doing something silly along the way, but I can't find my mistake...

Homework Statement


Find the energy stored in a uniformly charged solid sphere of radius R and charge q. Do it three different ways:
(a) Use Eq. 2.43
(b) Use Eq. 2.45
(c) Use Eq. 2.44

Homework Equations


Eq. 2.43 [tex]W=\frac{1}{2}\int \rho V d\tau[/tex]
Eq. 2.45 [tex]W=\frac{\epsilon_{0}}{2}\int_{all space} E^2 d\tau[/tex]
Eq. 2.44 [tex]W=\frac{\epsilon_{0}}{2}\left(\int_{V} E^2 d\tau + \oint_{S} VE\cdot da\right)[/tex]

The Attempt at a Solution



For part (a) we have:
[tex]E(r)=\left\{\begin{array}{cc}\frac{kq}{r^2}, &\mbox{ if } r > R\\ \frac{kq}{R^3} r, &\mbox{ if } r < R \end{array}\right<br /> \\ \Rightarrow V(r) = \left\{\begin{array}{cc}\frac{kq}{r}, & r > R\\ \frac{3kq}{2R} - \frac{kq}{R^3} r^2, & r < R \end{array}\right[/tex]

Since it's uniformly charged, we know that:
[tex]\rho = \frac{3q}{4\pi R^3} \\[/tex]

Then we can evaluate Eq. 2.43 to get:
[tex]W=2\pi\rho\int_{0}^R V(r) r^2 dr \\<br /> = \frac{9}{20} \frac{kq^2}{R}[/tex]

For (b) and (c) however I get:

[tex]W=\frac{3}{5} \frac{kq^2}{R}[/tex]

So they are off by a factor of 3/4.

If you need to see more work to help me find my mistake, let me know and I will post more details.
 
Last edited:
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E is the NEGATIVE gradient of the potential. You have a sign error. Also as written V is not even continuous across r=R, but I assume that is just a typo.
 
Thanks, there's no sign errors, but I forgot a factor of 1/2 in the second term of V for r<R
 
NeoDevin said:
Thanks, there's no sign errors, but I forgot a factor of 1/2 in the second term of V for r<R

I guess you're right. The sign error was mine!
 
Yeah, I worked it out with the correct expression for V, and it gives me the same answer now, thanks.
 
NeoDevin, you wouldn't happen to be in professor schick's class would you?
 

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