# Electric Potential energy for geometric objects

1. Sep 11, 2009

### simpleton

I recently found out how to calculate the potential energy of a charged insulated sphere of radius R and charge density rho. I would like to know how to calculate the potential energy for other geometric objects, such as a line of charges, a sheet of charges, or a pyramid of charges. I don't really see any symmetry that can be exploited, so I hope you guys can give me some help. Thanks :)

2. Sep 11, 2009

### Staff: Mentor

Simply integrate Coulomb's law over any charge distribution to get the field. The only subtlety is that due to the spherical symmetry you were able to do a scalar integration instead of a vector integration, but that is not true in general.

3. Sep 11, 2009

### simpleton

I don't see how you can do that :(. Can you give me some guidelines, or maybe show me how using the easiest example (other than the sphere)?

4. Sep 11, 2009

### Staff: Mentor

Say you have a completely arbitrary charge distribution and that the charge density is given (in Cartesian coordinates) by the function $\rho (X,Y,Z)$ then the E field at some point in space is given by:

$$E(x,y,z)=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty } \rho (X,Y,Z) k_e \frac{ (x-X,y-Y,z-Z)}{\left((x-X)^2+(y-Y)^2+(z-Z)^2\right)^{3/2}}dXdYdZ$$

5. Sep 11, 2009

### Staff: Mentor

Oops, thanks for the PM. I was giving you the answer for the E-field and you were asking about the potential energy (I assume you mean potential or voltage rather than potential energy). Luckily the potential is a scalar field so it is easier to calculate:

$$V(x,y,z)=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\frac{\rho (X,Y,Z) k_e }{\sqrt{(x-X)^2+(y-Y)^2+(z-Z)^2}}dXdYdZ$$

PS Unfortunately I do not know of a simpler equation that works in general. Everything else that is simpler requires some symmetry.

Last edited: Sep 11, 2009
6. Sep 11, 2009

### simpleton

Sorry, I still don't understand. May I know why charge density is a function and not a constant? Also, what is the difference between capital X and small x, and similar for y and z? I also don't understand why you are integrating from negative infinity to positive infinity, and I am not sure how finding the potential (potential between what?) will help. I guess you use U = qV, where q is the total net charge in the object. But that does not really make sense to me :S.

My calculus is not that good. I can do maths integration rather okay, but I cannot look at the integral and picture what is going on easily. Can you please explain it? Thanks a lot.

7. Sep 11, 2009

### Staff: Mentor

Charge density is a function and not a constant because the charge density changes as a function of position. For example, if you have a spherical charge then the charge density may be 1 for positions inside the sphere and 0 outside the sphere.

The integration ranges over infinity because, in general, your charge can be non-zero anywhere.

X and x are both coordinates in the x direction. The only difference is that you are integrating over X. After the integration all of the X, Y, Z terms are gone and you are only left with the voltage as a function of x, y, and z.

8. Sep 12, 2009

### simpleton

Is ke the the constant 1/(4*pi*epilson-nought)?

Also, I am still not too sure of the difference between x and X :S. What do you mean when you say you integrate over X?

Can you show me how you use this formula to calculate the potential energy of an easy example? Say:

The potential energy of a charged sphere of charge density 5 Cm^-3 and radius 1m.

EDIT: Just to clarify, for the example above, does rho (x, y, z) = 5 if point (x, y, z) is within the sphere, and 0 if point (x, y, z) is outside the sphere?

9. Sep 12, 2009

### ZapperZ

Staff Emeritus
You need to be a bit reasonable here. You are asking for a lesson in E&M that is typically taught in advanced undergraduate physics classes that covers several weeks.

I'm sure DaleSpam has more patience and has a high willingness to continue with this. But it would be so much easier if you have an undergrad E&M text with you and work through the first few chapters on electrostatics. Many of the questions you're asking are basic, fundamental questions that are clearly and sufficiently addressed in such text (eg. Griffiths' excellent E&M book). Furthermore, this is where using diagram is not only useful, but almost a necessity. It would be so much easier to address specific question that you have upon working on such material, rather than give broad lessons.

Zz.

10. Sep 12, 2009

### Staff: Mentor

Yes.

Let's say that you are interested in the potential at some specific point (x,y,z) = (1,2,3). So you plug in those numbers into the equation above. Now you note that you still have the variables X Y and Z to integrate over. This is because the potential at point (1,2,3) is affected not just by the charge distribution at (1,2,3), but by charges located anywhere! Therefore you have to integrate over all space in order to determine the potential at any single point. So you integrate over X Y and Z and then finally you have the potential at (1,2,3). If you want to find the potential at some other point then you have to repeat the integral there.

Yes.

Since the charge density is 0 outside the sphere we can do this by changing the limits of integration appropriately:

$$\int _{-1}^1\int _{-\sqrt{1-Z^2}}^{\sqrt{1-Z^2}}\int _{-\sqrt{-Y^2-Z^2+1}}^{\sqrt{-Y^2-Z^2+1}}\frac{\rho (X,Y,Z) k_e}{\sqrt{(x-X)^2+(y-Y)^2+(z-Z)^2}}dXdYdZ$$

Since the charge density and ke are constant within those limits we can pull them out and get:

$$\rho k_e \int _{-1}^1\int _{-\sqrt{1-Z^2}}^{\sqrt{1-Z^2}}\int _{-\sqrt{-Y^2-Z^2+1}}^{\sqrt{-Y^2-Z^2+1}}\frac{1}{\sqrt{(x-X)^2+(y-Y )^2+(z-Z)^2}}dXdYdZ$$

Which you can evaluate numerically at any point (x,y,z) to get the potential there.

11. Sep 14, 2009

### simpleton

Ok, thanks a lot :)

12. Sep 19, 2009

### simpleton

Hi,

I was told another formula to calculate energy, 0.5 * E^2 * e0. This is derived from the capacitors equations, but supposedly this is a general equation.

Anyway, I tried to use this equation to calculate the energy of a sphere. Using the integration by shells method, I have gotten the answer:

(4 * rho^2 * pi * R^5) / (15 * e0)

where rho is the charge density and R is the radius.

I used the electric field equation and see whether I can get the same answer. Firstly, I used Gauss Law to find the electric field. If I take an imaginary surface that has a radius v > R, then I get:

E * (4 * pi * v^2) = (rho * 4 * pi * r^3) / (3 * e0)

E = (rho * r^3) / (3 * v^2 * e0)

0.5 * E^2 * e0 = (r^6 * rho^2) / (18 * v^4 * e0)

Then I integrate this all over space. Using circular coordinates, I chuck in the term v^2 * sin theta d(thetda) d(phi) d(r) and integrated r to infinity to R, theta from pi to 0, and phi from 2*pi to 0. This give me:

Integral (r^6 * rho^2 * sin(theta) ) / (18 * v^2 * e0)

= (r^5 * rho^2 * pi) / (9 * e0)

However, this answer is not complete, because I have to calculate the electric field within the sphere itself too. Therefore, I used Gauss Law again, and now the radius of the imaginary surface is the same as the radius of the sphere it has enclosed:

E * (4 * pi * r^2) = (rho * 4 * pi * r^3) / (3 * e0)

E = (r * rho) / (3 * e0)

0.5 * E^2 * e0 = (r^2 * rho^2) / (18 * e0)

Then, I chuck in the term r^2 sin(theta) dr d(theta) d(phi) and integrate r from R to 0, theta from pi to 0, phi from 2 * pi to 0, and I got the answer:

(2 * pi * rho^2 * r^5) / (45 * e0)

So, the total energy I get is:

(2 * pi * rho^2 * r^5) / (45 * e0) + (r^5 * rho^2 * pi) / (9 * e0)

= (7 * pi *rho^2 * r^5) / (45 * e0)

This answer is different from (4 * rho^2 * pi * R^5) / (15 * e0). Can someone tell me whether I did any step wrongly?

Thanks.

13. Sep 21, 2009

### Staff: Mentor

The equation I gave you was for calculating the potential (voltage) at a point. The energy formula from capacitors that you gave calculates the energy density of the field itself at a point (which then you integrate over the volume to get then energy). They are equations for different things, not just different equations for the same thing.

14. Sep 22, 2009

### simpleton

Sorry, as my integration skills are not up to mark, I was unable to use the equations you gave to solve the problems. I will try to solve the problems using them when I have a better grasp of integration.

As for the my wrong answer, I managed to get the correct answer after trying it some more times. It turns out that one of my integrations are done wrongly :(.