Work done by moving a point charge

[Jake 7174]

Homework Statement

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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 W_AB = qV_AB. First let us assume that the side of the triangle is a and later we can substitute a number for it.

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

Homework Equations

V = q / (4πε₀r)

W = qV_AB

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. Let's make an equilateral triangle with a charge q₁ at the origin, q₂ at (4,0), and q₃ at (2,2sqrt(3))

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

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

Then V_AB = V_B - V_A = 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.

 

[CrazyNinja]

Are you sure the answer is 72mJ?

 

[Jake 7174]

Are you sure the answer is 72mJ?

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

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

Is this not correct?

 

[CrazyNinja]

Okay the answer is correct. I made a small mistake. Now, what is your doubt?

 

[Jake 7174]

Okay the answer is correct. I made a small mistake. Now, what is your doubt?

It is the setting up of the equation for V_AB. 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 V_A and V_B, 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.

 

[CrazyNinja]

I essentially treated a charge as a test charge to find VA and VB, but shouldn't this charge contribute to V as well.

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.

 

[Jake 7174]

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.

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

 

[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.