A point charge q1 = +2.80 µC is at origin. How far should the second point charge of +5.20 µC be placed to have electric potential energy of 0.600 J?
Answer
The distance is
"r=\\frac{kQQ'}{U}\\\\=\\frac{9*10^9*2.8*5.2*10^{-12}}{0.6}\\\\=6.5cm"
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