Electric Potential/Equilateral Triangle

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The forum discussion centers on calculating the work required to move a charge in an equilateral triangle configuration with sides of 0.5 mm and equal charges of 4 pC. The user initially calculated the electric potential at the midpoint between two charges as 288 V, leading to a work calculation of 1152 pJ. However, the correct answer is 576 pJ, which the user arrived at after realizing the need to account for the potential difference from the charge's initial position rather than from infinity. This adjustment clarified the misunderstanding in the work calculation.

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Siirous
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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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