# Homework Help: Electric potential energies

1. Apr 22, 2012

### tracyellen

Rank the electric potential energies of the systems of charges shown in the figure below from largest to smallest. Indicate equalities if appropriate. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)

I was thinking A=C>B=D

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2. Apr 22, 2012

### SammyS

Staff Emeritus
Hello tracyellen. Welcome to PF !

A & C have higher Potential Energies than B & D. Otherwise, you're incorrect.

3. Apr 23, 2012

### tracyellen

I was thinking that because q is the same distance that A and C/B and D would equal eachother because they have the same potnetial energy. If this is not the case due to the shape of each I would say that A>C>D>B due to the triangle being smaller and having a greater potential energy? Thank you so much for your help!

4. Apr 23, 2012

### SammyS

Staff Emeritus
That's not the right order. That analysis is overly simple .

How do you calculate the potential energy of such a system?

5. Apr 23, 2012

### tracyellen

I am sorry. I don't really know how to find a PE of an item without any numbers. I am used to a length or something to help me out. That is why I am having such a hard time with this problem. :(

6. Apr 23, 2012

### tracyellen

Maybe E=KEo....but again I don't know how to make that work with this problem. I am not really looking for an answer for I have already turned in the wrong one. I am just trying to figure out how I got it wrong and how to do it right. :)

7. Apr 23, 2012

### SammyS

Staff Emeritus
I think it's helpful to show the figure full size.

It takes zero work to bring the first positive charge into position (for all four configurations).

For A & C, bring a second positive charge into position into position a distance, d, from the first charge. The potential energy stored in such a system is equal to the amount of work required to bring the second charge into position. Call this amount of energy, E0 .
For A: Bringing a third positive requires twice the work, i.e. 2E0. Total PE is 3E0.

For C: Bringing a third positive charge requires a little less work than for A. E0 to overcome the repulsion of the closer charge. A bit less to overcome the repulsion of the other charge. But then there is a fourth charge!
It takes 2E0 to overcome the repulsion of the two closest charges. A bit less to overcome the repulsion of the other charge.​
The total for C: (4 + _)E0 .

For B & D, bring a second charge, this time a negative charge, into position into position a distance, d, from the first charge. The work needed to do this is -E0, so that's the potential energy stored in such a system.
For B: Bringing a third charge into position requires zero work. E0 to

.

8. Apr 23, 2012

### SammyS

Staff Emeritus
I think it's helpful to show the figure full size.

It takes zero work to bring the first positive charge into position (for all four configurations).

For A & C, bring a second positive charge into position into position a distance, d, from the first charge. The potential energy stored in such a system is equal to the amount of work required to bring the second charge into position. Call this amount of energy, E0 .
For A: Bringing a third positive requires twice the work, i.e. 2E0. Total PE is 3E0.

For C: Bringing a third positive charge requires a little less work than for A. E0 to overcome the repulsion of the closer charge. A bit less to overcome the repulsion of the other charge. But then there is a fourth charge!
It takes 2E0 to overcome the repulsion of the two closest charges. A bit less to overcome the repulsion of the other charge.​
The total for C: (4 + _)E0 .

For B & D, bring a second charge, this time a negative charge, into position into position a distance, d, from the first charge. The work needed to do this is -E0, so that's the potential energy stored in such a system.
For B: Bringing a third charge, this one positive, into its position requires zero work. E0 to overcome the repulsion of the 1st positive charge & -E0 due to the attraction of the negative charge.
Total PE for B: -E0 .​
D is a little more involved.

Bringing a third charge, this the other positive one a distance, d, from the negative charge, requires -E0 due to the attraction of the negative charge plus some positive work, a bit less than E0 overcome the repulsion of the 1st positive charge, because the separation is greater than d. Bringing the forth charge to its position requires 2E0 to overcome the repulsion of the other two positive charges plus some negative work due to the attraction of the negative charge. This negative PE is a bit smaller in magnitude than E0 and in fact cancels the the previous positive PE which was a bit less than E0.
Total PE for D: 0 .​

9. May 6, 2012

### tracyellen

Great thank you. This makes more sense. I appreciate your help!!