Chapter 3: Problem 147
Use Coulomb's law, $$V=\frac{Q_{1} Q_{2}}{4 \pi \epsilon_{0} r}=2.31 \times 10^{-19} \mathrm{J} \cdot \mathrm{nm}\left(\frac{Q_{1} Q_{2}}{r}\right)$$ to calculate the energy of interaction, \(V\), for the following two arrangements of charges, each having a magnitude equal to the electron charge.
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