Ranking Electric Potential Energies: A=C>B=D

[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

Please help me if I am wrong!

 

[SammyS]

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

Please help me if I am wrong!

Hello tracyellen. Welcome to PF !

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

Explain your thinking, so we can help you.

 

[tracyellen]

I was thinking that because q is the same distance that A and C/B and D would equal each other 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!

 

[SammyS]

I was thinking that because q is the same distance that A and C/B and D would equal each other because they have the same potential 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!

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

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

 

[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. :(

 

[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. :)

 

[SammyS]

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. :(

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. :)

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

Start with no charge -- empty space.

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

​

.

 

[SammyS]

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. :(

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. :)

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

Start with no charge -- empty space.

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 .​

 

[tracyellen]

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