Work done assembling a system of charges?

In summary, the concept of electrostatic energy stored in a collection of static point charges has caused confusion for the speaker. They quote an article and provide an example to illustrate their confusion, as the work done to assemble the charges is negative despite the electric force handling their attraction. The speaker is unsure why this is the case, as it goes against the known formula for work done in neutral bodies.
  • #1
256
18
This is a concept that has long since plagued me. I will quote an article off of Google about the concept I am confused about:

"Consider a collection of
img532.png
static point charges
img561.png
located at position vectors
img562.png
(where
img536.png
runs from 1 to
img532.png
). What is the electrostatic energy stored in such a collection? Another way of asking this is, how much work would we have to do in order to assemble the charges, starting from an initial state in which they are all at rest and very widely separated? " [http://farside.ph.utexas.edu/teaching/em/lectures/node56.html]

So, I understand the general concept. But I am confused in one of the cases. The case where the work that we do is negative (i.e. the work that we have to do to assemble the charge is < 0).

Here is an example that I will use to illustrate my confusion:

Say that we have two charges q in space, one positive, and one negative, and want to move them close to each other, to a distance of r.

The answerof course, is that the work that we need to do to bring the negative charge to the positive charge is then W = -kq^2/r < 0. I know this is the answer, and I know how to solve these problems. But I have no idea why this is true.

I am confused about the concept. I will try to explain what I am confused about below:

It is known that W = Fd. In this case , d = (infinity - r)

But I am confused about this, because essentially the charges attract one another. Therefore, we don't really have to apply any force to make them get close to each other, because an electric force will handle this for us. So our applied force F = 0. Then W = 0*d = 0. But of course, this is wrong. Because we know the answer to be W = -kq^2/r. But why?
 
Last edited by a moderator:
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  • #2
x86 said:
This is a concept that has long since plagued me. I will quote an article off of Google about the concept I am confused about:

"Consider a collection of
img532.png
static point charges
img561.png
located at position vectors
img562.png
(where
img536.png
runs from 1 to
img532.png
). What is the electrostatic energy stored in such a collection? Another way of asking this is, how much work would we have to do in order to assemble the charges, starting from an initial state in which they are all at rest and very widely separated? " [http://farside.ph.utexas.edu/teaching/em/lectures/node56.html]

So, I understand the general concept. But I am confused in one of the cases. The case where the work that we do is negative (i.e. the work that we have to do to assemble the charge is < 0).

Here is an example that I will use to illustrate my confusion:

Say that we have two charges q in space, one positive, and one negative, and want to move them close to each other, to a distance of r.

The answerof course, is that the work that we need to do to bring the negative charge to the positive charge is then W = -kq^2/r < 0. I know this is the answer, and I know how to solve these problems. But I have no idea why this is true.

I am confused about the concept. I will try to explain what I am confused about below:

It is known that W = Fd. In this case , d = (infinity - r)

But I am confused about this, because essentially the charges attract one another. Therefore, we don't really have to apply any force to make them get close to each other, because an electric force will handle this for us. So our applied force F = 0. Then W = 0*d = 0. But of course, this is wrong. Because we know the answer to be W = -kq^2/r. But why?
x86 said:
This is a concept that has long since plagued me. I will quote an article off of Google about the concept I am confused about:

"Consider a collection of
img532.png
static point charges
img561.png
located at position vectors
img562.png
(where
img536.png
runs from 1 to
img532.png
). What is the electrostatic energy stored in such a collection? Another way of asking this is, how much work would we have to do in order to assemble the charges, starting from an initial state in which they are all at rest and very widely separated? " [http://farside.ph.utexas.edu/teaching/em/lectures/node56.html]

So, I understand the general concept. But I am confused in one of the cases. The case where the work that we do is negative (i.e. the work that we have to do to assemble the charge is < 0).

Here is an example that I will use to illustrate my confusion:

Say that we have two charges q in space, one positive, and one negative, and want to move them close to each other, to a distance of r.

The answerof course, is that the work that we need to do to bring the negative charge to the positive charge is then W = -kq^2/r < 0. I know this is the answer, and I know how to solve these problems. But I have no idea why this is true.

I am confused about the concept. I will try to explain what I am confused about below:

It is known that W = Fd. In this case , d = (infinity - r)

But I am confused about this, because essentially the charges attract one another. Therefore, we don't really have to apply any force to make them get close to each other, because an electric force will handle this for us. So our applied force F = 0. Then W = 0*d = 0. But of course, this is wrong. Because we know the answer to be W = -kq^2/r. But why?
It is known that W=F.d for neutral bodies but the case changes totally for charged bodies. We have equation of force F=qE for charged bodies so we can not put F=0 in case of charged bodies.
 
Last edited by a moderator:
  • #3
That the work required to assemble the charges is negative simply means that you would extract work from doing so. Unless you counter the electric force with a force in the other direction, the final configuration will involve moving charges, which simply means the work done by the electric force has instead gone into kinetic energy of the constituents.
 
  • #4
Muhammad Nauman said:
It is known that W=F.d for neutral bodies but the case changes totally for charged bodies.

There is absolutely nothing special about neutral bodies in the work-energy theorem. The only relevant issue is which force you consider and the path taken, which will determine if the force you considered did work or not.
 
  • #5
Orodruin said:
That the work required to assemble the charges is negative simply means that you would extract work from doing so. Unless you counter the electric force with a force in the other direction, the final configuration will involve moving charges, which simply means the work done by the electric force has instead gone into kinetic energy of the constituents.

Ah, so by doing negative work I'm essentially slowing the particles to a stop, at this snapshot in time?
 

Related to Work done assembling a system of charges?

1. What is work done assembling a system of charges?

The work done assembling a system of charges refers to the energy required to bring together a group of charges from an infinite distance apart to their final positions in a given configuration.

2. How is work done assembling a system of charges calculated?

The work done assembling a system of charges is calculated by multiplying the total charge of the system by the potential difference between the initial and final positions of the charges. This can be expressed mathematically as W = qΔV.

3. What is the significance of work done assembling a system of charges?

The work done assembling a system of charges is important because it represents the amount of energy that has been stored in the system. This energy can be released later on, potentially causing changes in the system's behavior.

4. How does the distance between charges affect the work done assembling a system of charges?

The work done assembling a system of charges is directly proportional to the distance between the charges. This means that as the distance between charges increases, the work done also increases, and vice versa.

5. Can the work done assembling a system of charges be negative?

Yes, the work done assembling a system of charges can be negative if the charges are of opposite signs and are brought closer together. This means that work is being done by the system, and the charges are losing potential energy as they move closer together.

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