Electric Potential/Equilateral Triangle

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The discussion revolves around calculating the work needed to move a charge in an equilateral triangle configuration with equal charges. The user initially calculated the electric potential at the midpoint between two charges but arrived at an incorrect total work value. The key correction involved recognizing that the charge was not being moved from infinity, allowing for the subtraction of the potential energy already present at its initial position. After adjusting the calculations, the user successfully arrived at the correct answer of 576 pJ. The importance of considering the initial position in potential energy calculations was highlighted as a crucial step in solving the problem.
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I'm stuck on this one, which I thought would be simple, hoping for a little nudge in the right direction.

The problem states an equilateral triangle, sides of .5mm (which I converted to .5E-3m), all equal charges of 4pC.

How much work must be done to move one charge to a point equidistant from the other two and on the line that joins them.

So here's how I went about figuring out how much work to move one point in between the other two. I set the triangle up with two points on the x-axis, .25E-3 meters from the origin, and the third point on the +ve y axis.

I said V=kq/r = (9E9)(4E-12)/.25E-3 = 144V from one charge on the x-axis, then multiplied by two, since the other charge is mirrored. So 288 Volts.

I then tried to move the charge on the yaxis into the origin by saying
U = qV so U = (4E-12)(288) =1152E-12 J = 1152 pJ

However, their answer is 576 pJ, half of what I got... where am I going wrong?

Thanks in advance,
Rob
 
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Think about the potential difference between where the charge was initially placed and its final position.
 
Ah ha, because I'm not bringing it in from infinity I can subtract the joules needed to bring it to it's current position, then leftover will be what I need, it works! Got the answer! Thanks a lot.
 
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