# A system of point charges

1. Jun 3, 2013

### carnot cycle

1. The problem statement, all variables and given/known data
Two point charges are located on the x-axis, q1 = -e at x = 0 and q2=+e at x=a.(a)Find the work that must be done by an external force to bring a third point charge q3 = +e from infinity to x = 2a. (b) Find the total potential energy of the system of three charges.

3. The attempt at a solution

I am struggling with part a of this question

I found that the net force acting on q3 at x = 2a due to the charges q1 and q2 is the following:

F = 3Ke2/4a2

Where K is Coulomb's constant. This was done using Coulomb's law.

Now to find the work done by an external force in bringing the charge q3 from infinity to x = 2a, can't I just find the work done by the electric force in pushing the charge q3 from x = 2a to infinity? Should those two be equal in magnitude? If so, then I obtained the following answer after integration:

W = 3Ke2/(8a)

However, this answer is incorrect. The book states the following: W = 4Ke2/(8a). I am wondering where I went wrong. Thanks.

2. Jun 3, 2013

### Simon Bridge

The work to get from a to b is the difference in potential energy at those points.

3. Jun 3, 2013

### rude man

Yes.
Well, you didn't integrate the two forces over the distances a → ∞ and 2a → ∞ correctly.

An easier way is to just compute the potential at 2a = V(2a). That requires no integration. V(2a) is the work done to bring a UNIT charge from ∞ to 2a. So the work needed to bring a charge of e to that same spot is
eV(2a).

4. Jun 4, 2013

### carnot cycle

Why is it incorrect to find the net force acting on q3 (from Coulomb's law) and integrating that force from 2a to infinity?

Last edited: Jun 4, 2013
5. Jun 4, 2013

### rude man

It's not. I said your integration was not performed correctly.

You have two integrals to perform: one from ∞ to a and one from ∞ to 2a, right?

Last edited: Jun 4, 2013
6. Jun 4, 2013

### carnot cycle

I integrated the force which I found from Coulomb's Law

∫3Ke2/4a2 * da

This was done from x = 2a to infinity. I pulled out the 3Ke2/4 factor since it is a constant and was left with this.

3Ke2/4 ∫1/a2 * da

The anti-derivative of 1/a2 is -1/a which was evaluated from an upper limit of infinity to a lower limit of x = 2a.

(-1/∞)-(-1/2a) = 1/2a

so then I multiply that with the constant term to obtain 3Ke2/8a but that is incorrect.

I don't see why I have two integrals to find the work done by the electric force in moving q3 from 2a to infinity :/

7. Jun 4, 2013

### rude man

Because there are two forces acting on your third charge.

One is from -e located at x = 0. So you need to integrate that force from infinity to (2a - 0) = 2a.

The other is from the +e located at x = a, so that force needs to be integrated from infinity to
(2a - a) =a.

Since the +e charge is located closer to 2a than the -e charge you know ahead of time that the total answer will be positive (positive work done ON the third charge).

8. Jun 4, 2013

### carnot cycle

But don't both charges create a net force that pushes q3 from x = 2a to x = infinity? So shouldn't the limits of integration be infinity to 2a? Sorry that I keep asking the same question. I just can't conceptually grasp it

9. Jun 4, 2013

### rude man

Take a point x > 2a. The force from the -e charge at that point is ke^2/x^2. The force from the +e charge is ke^2/(x-a)^2. They're different force magnitudes and directions.

OK, you could combine them into one net force F_net = ke^2{1/x^2 - 1/(x-a)^2} and integrate that from infinity to 2a. Same answer.

Remember, this is the force applied TO the third charge to bring it from x = infinity to x = 2a. The force ON the third charge by the first two charges is the negative of F_net. Your integral must be the integral of force applied TO the third charge.

10. Jun 4, 2013

### carnot cycle

So the force from the -e charge is ke^2/4a^2 and the force from the +e charge is ke^2 /a^2?

and I subtract these two to find that the net force is 3ke^2/4a^2, which I integrate from x= infinity to x = 2a? If so, then i performed these steps in post 6 but did not retrieve the right answer.

11. Jun 5, 2013

### rude man

You are stating the forces at x = 2a only.

Your integral, as I have said, is from x = ∞ to x = 2a. Study my last post again.

12. Jun 5, 2013

### Simon Bridge

Only at point x=2a ... it does not have that relationship for any other point.

... initially you got the limits the wrong way around, but you fixed that, now what you are saying is that you did this:
$$W=\int_\infty^{2a}F(2a)da$$ ... see the problem?

By the definition of work: the work done moving from point $x$ to $x+dx$ is $F(x)dx$. To go from $x=\alpha$ to $x=\beta$ you have to integrate over all the values of $x$ in between $\alpha$ and $\beta$.

What you said you did was $$W=\int_\alpha ^\beta F(\beta )d\beta$$ when you should have done: $$W=\int_\alpha ^\beta F(x)dx$$ ...i.e. you made the substitution too soon.
Leave the equation for the forces as functions of x and then do the integral.

13. Jun 6, 2013

### carnot cycle

Ohhhh okay, now I see where I went wrong. Thanks so much for helping!

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