91Ó°ÊÓ

For both arrangements, we have the same charges, with both charges having a magnitude equal to the electron charge (\(Q_1 = Q_2 = e\)), where - Electron charge \(e\) : \(1.6 \times 10^{-19} C\) - Vacuum permittivity constant \(\varepsilon_0\) : \(8.85 \times 10^{-12} C^2 / Nm^2\) For each arrangement, we also need the distance between the charges, \(r\). For this exercise, we are not given any specific values for 'r' in the arrangements. Therefore, we will treat 'r' as a variable and calculate the interaction energy in terms of 'r' for both arrangements.

To calculate the energy of interaction, we can plug the known variables into Coulomb's Law formula: \[V = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 r} = 2.31 \times 10^{-19} \cdot J \cdot nm \left(\frac{Q_1 Q_2}{r} \right)\] We can substitute the values of \(Q_1 = Q_2 = e\) and \(\varepsilon_0\): \[V = \frac{e^2}{4 \pi \varepsilon_0 r} = 2.31 \times 10^{-19} \cdot J \cdot nm \left(\frac{e^2}{r} \right)\] Now, let's plug in the value of \(e = 1.6 \times 10^{-19} C\): \[V = \frac{(1.6 \times 10^{-19})^2}{4 \pi (8.85 \times 10^{-12}) r} = 2.31 \times 10^{-19} \cdot J \cdot nm \left(\frac{(1.6 \times 10^{-19})^2}{r} \right)\] Now, we can simplify: \[V = \frac{(2.56 \times 10^{-38})}{4 \pi (8.85 \times 10^{-12}) r} = 2.31 \times 10^{-19} \cdot J \cdot nm \left(\frac{2.56 \times 10^{-38}}{r} \right)\] Therefore, the interaction energy for both arrangements (in terms of 'r') is: \[V = 2.31 \times 10^{-19} \cdot J \cdot nm \left(\frac{2.56 \times 10^{-38}}{r} \right)\]