A point charge q1 = +3.90 is at origin. How far should the second point charge of +6.30 be placed to have electric potential energy of 0.700 J?
Answer
The distance must
"R=\\frac{Kqq'}{U}\\\\=\\frac{9*10^9*3.9*6.30}{0.7}\\\\=315.9m"
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