Electrostatics, Energy of a uniformly charged sphere

In summary, neoDevin gets the same answer for parts b and c, but not for a. He is uncertain of what he is doing wrong.
  • #1
NeoDevin
334
2
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
\\ \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 \\
= \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:
Physics news on Phys.org
  • #2
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.
 
  • #3
Thanks, there's no sign errors, but I forgot a factor of 1/2 in the second term of V for r<R
 
  • #4
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!
 
  • #5
Yeah, I worked it out with the correct expression for V, and it gives me the same answer now, thanks.
 
  • #6
NeoDevin, you wouldn't happen to be in professor schick's class would you?
 

1. What is electrostatics?

Electrostatics is the study of stationary electric charges and the forces they exert on each other. It is a branch of physics that deals with the behavior of electric charges at rest.

2. What is the energy of a uniformly charged sphere?

The energy of a uniformly charged sphere is the sum of all the potential energies of the individual charges that make up the sphere. It can be calculated using the equation: E = (3/5)*((k*q^2)/r), where k is the Coulomb constant, q is the charge of the sphere, and r is the radius of the sphere.

3. How is the energy of a uniformly charged sphere related to its charge and radius?

The energy of a uniformly charged sphere is directly proportional to the square of its charge and inversely proportional to its radius. This means that as the charge of the sphere increases, its energy also increases, and as the radius of the sphere increases, its energy decreases.

4. Can the energy of a uniformly charged sphere be negative?

Yes, the energy of a uniformly charged sphere can be negative. This occurs when the sphere is negatively charged and there are more positively charged particles in its surroundings, causing the sphere to have a lower potential energy.

5. How is the energy of a uniformly charged sphere affected by the presence of other charged objects?

The energy of a uniformly charged sphere can be affected by the presence of other charged objects through electrostatic interactions. If the other object has a different charge than the sphere, it can either attract or repel the sphere, changing its energy. Additionally, the distance between the objects also plays a role in the energy of the sphere, as the closer the objects are, the stronger their interaction and the greater the effect on the sphere's energy.

Similar threads

  • Advanced Physics Homework Help
Replies
19
Views
700
Replies
11
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
300
  • Advanced Physics Homework Help
Replies
13
Views
2K
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
893
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
6
Views
2K
Back
Top