# Work done by moving a point charge

1. Feb 14, 2016

### Jake 7174

1. The problem statement, all variables and given/known data

Three point charges each of 4 µC are situated at the three corners of an equilateral triangle of side 4 m. Find the work done in moving one of them to a point mid-way between the other two.

Solution: The Potential diﬀerence VAB between two points A and B is the work done in moving a unit positive charge from A to B. The work done in moving a test charge q from A to B is WAB = qVAB. First let us assume that the side of the triangle is a and later we can substitute a number for it.

VAB = 2 * [q / (4πε0) * (2/a −1/a)] = 2q / (4πε0a) = 72 / s kV

2. Relevant equations

V = q / (4πε0r)

W = qVAB

3. The attempt at a solution

I have the solution part of it is shown above. The answer is 72 mJ. I think I understand it but am having trouble convincing myself with regaurds to portion of the solution shown above. I need some conceptual assistance. Here is what I am seeing. Lets make an equilateral triangle with a charge q1 at the origin, q2 at (4,0), and q3 at (2,2sqrt(3))

I belive VA at q3 = 2 * q / (4πε0a) where a is the radius (defined in the solution)

I belive VB at (2,0) when moving q3 down = 2 * q / (4πε0(a/2)) = q / (πε0a)

Then VAB = VB - VA = q / (πε0a) - 2 * q / (4πε0a) = 2q / (4πε0a)

This is the solution I am looking for but I want to make sure my reasoning is sound. My equation is equivalent to but is not set up like the one in the solution. The way the solution shows it in step 1 seems like an odd way to express it.

I am also having trouble convincing myself of this result because q3 is a point charge and not just some test point. I want ta say that there must be a voltage created by q3 and that my math has neglected to account for it.

Any help is greatly appreciated. It is exam day tomorrow and I don't want to blow it on lack of conceptual understanding.

2. Feb 14, 2016

### CrazyNinja

Are you sure the answer is 72mJ?

3. Feb 14, 2016

### Jake 7174

If it is not; then the solution given by my professor is incorrect. VAB ends up being 18 kV according to the solution.

W = qVAB = 4*10^(-6) * 18000 = .072 J

Is this not correct?

Last edited: Feb 14, 2016
4. Feb 14, 2016

### CrazyNinja

5. Feb 14, 2016

### Jake 7174

It is the setting up of the equation for VAB. Is the method I used to find it good or did I get lucky? I essentially treated a charge as a test charge to find VA and VB, but shouldn't this charge contribute to V as well. After all it has charge equal to the other two.

The math works out so I know I am a bit mixed up conceptually. I trust the numbers more than myself.

6. Feb 14, 2016

### CrazyNinja

A charge does not contribute to the field which acts on it. Clearly speaking, it does not exert a force on itself. Hence, there is no potential assosciated with itself.

To avoid confusion, do it by using potential energy instead of setting up the potential as you did. Try it and tell me if you feel comfortable that way. If not, we will try another alternative. Take initial potential energy, final potential energy and get the difference.

7. Feb 14, 2016

### Jake 7174

I think I am good. I feel confident. Thank you

8. Feb 14, 2016

### CrazyNinja

Write the exam well. Try a few more problems like this using the potential energy method. Get the theory right, the questions will automatically get answered.