Calculating Work and Potential Energy in a System of Point Charges

[carnot cycle]

Homework Statement

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.

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.

 

[Simon Bridge]

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

 

[rude man]

Homework Statement

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.

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?

Yes.

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.

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).

 

[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?

 

[rude man]

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

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?

 

[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 :/

 

[rude man]

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 :/

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).

 

[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

 

[rude man]

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

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.

 

[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.

 

[rude man]

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.

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.

 

[Simon Bridge]

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.

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

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

find that the net force is 3ke^2/4a^2, which I integrate from x= infinity to x = 2a?

I integrated the force which I found from Coulomb's Law
∫3Ke2/4a2 * da ... This was done from x = 2a to infinity

... 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.

 

[carnot cycle]

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