Calculating Potential at a Point Due to Multiple Point Charges

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To calculate the electric potential at point A due to three point charges located at the corners of a square, the potential from each charge can be computed individually using the formula V = kQ/r. The confusion arises from understanding how to sum the potentials from multiple charges, especially when considering the reference point where V = 0 at a great distance. The correct approach involves adding the contributions from each charge, leading to the total potential expression. The final result can be simplified to match the textbook answer, confirming that the potential is indeed negative when considering the reference point at infinity. Understanding these concepts clarifies the calculation of potential in multi-charge systems.
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Three point charges are arranged at the corners of a square of side L. What is the potential at the fourth corner (point A), taking V = 0 at a great distance?

OK, I am very confused. First of all, Giancoli (my textbook) seems to explain how to do this for one point charge, but I am confused about doing this for three charges. Second, what do they mean by "at a great distance"? I am assuming I would want to use V = 1/4piE0 Q/r. So I am guessing that I would have to do some kind of vector deal? Help appreciated. :redface:
 
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Um never mind. So I thought about this again and decided to use V=kQ/L for each point individually -- does that make sense? And then add them up to get V total. So I wound up with ((1/square root 2) +1) x kQ/L. My book says the answer is square root 2 x k x Q/2L (square root 2 + 1). I believe this is the same thing?
 
That "taking V = 0 at a great distance" means that an infinity the potential is 0,which means that the potential is actually negative.So you'd have to add 3 negative #-s,that's all.

Daniel.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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