Change in energy between system of point charges

AI Thread Summary
The energy stored in a system of three identical point charges at the vertices of an equilateral triangle is 15 Joules. When one charge is moved to the midpoint of the opposite side, the new energy configuration is calculated to be 25 Joules. The difference in energy between the two configurations is determined to be 10 Joules. The initial calculations were confirmed to be correct, with some clarification on the mathematical setup. Overall, the discussion emphasizes understanding the relationship between charge distance and energy in point charge systems.
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Homework Statement


The energy stored by any pair of positive charges is inversely proportional to the distance between them, and directly proportional to their charges. Three identical point charges start at the vertices of an equilateral triangle, and this configuration stores 15 Joules of energy. How much more energy, in Joules, would be stored if one of these charges was moved to the midpoint of the opposite side?


Homework Equations





The Attempt at a Solution



We determine our formula to be $$U = \frac{kq^2}{r}.$$ Then from the given data, for three point charges we have $$U = 3 \frac{kq^2}{r} = 15J.$$ To compute the new position:

$$
U' = k(\frac{q^2}{r/2} + \frac{q^2}{r/2} + \frac{q^2}{r}) \\
U' = 5 \frac{kq^2}{r} \\
U' = 25J \\

$$

So the difference is $$U' - U = 10J.$$

I'm not quite sure if I am going about this the right way.
 
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I think you have gone about it almost the right way. I think the problem is where you did 1/2 + 1/2 + 1 = 5 This is not right.

PS welcome to physicsforums! :)

EDIT: I am tired and didn't read your post through properly. Sorry. Your maths was 2 + 2 + 1 = 5 which is of course correct. I think your answer is all OK, why do you think it is wrong?
 
Sorry, I think my typesetting make the fractions look confusing. I'm still learning to use Tex properly.
I just wanted to make sure that my set up was correct.
Thank you for your help and for the welcome!
 
yep, no worries :)
 
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