Answer to Question #138557 in Electric Circuits for Zase

Question

Two particles, with charges of 20.0 nC and 40.0 nC are placed at the points with coordinates (0, 4.00 cm) and (3.00 cm, 0). A particle with charge 10.0 nC is located at the origin. (a) Find the electric potential energy of the configuration of the three fixed charges.

Answer 1

According to https://en.wikipedia.org/wiki/Electric_potential_energy#Energy_stored_in_a_system_of_three_point_charges

the potential energy of the system can be calculated as

$U_E = k\left( \dfrac{q_1q_2}{r_{12}} +\dfrac{q_1q_3}{r_{13}} +\dfrac{q_2q_3}{r_{23}}\right).$

The distance between 1 and 2 charges is $r_{12} = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} = \sqrt{(0-3)^2+(4-0)^2} = 5\,\mathrm{cm}=0.05\,\mathrm{m}.$

The distance between 1 and 3 charges is $r_{13} = 4\,\mathrm{cm} = 0.04\,\mathrm{m}.$

The distance between 2 and 3 charges is $r_{23} = 3\,\mathrm{cm} = 0.03\,\mathrm{m}.$

So

$U_E = 9\cdot10^9\,\mathrm{Nm^2/C^2}\cdot\left( \dfrac{2\cdot10^{-8}\,\mathrm{C}\cdot4\cdot10^{-8}\,\mathrm{C}}{0.05\,\mathrm{m}} + \dfrac{2\cdot10^{-8}\,\mathrm{C}\cdot1\cdot10^{-8}\,\mathrm{C}}{0.04\,\mathrm{m}} + \dfrac{4\cdot10^{-8}\,\mathrm{C}\cdot1\cdot10^{-8}\,\mathrm{C}}{0.03\,\mathrm{m}}\right) = 3.09\cdot10^{-4}\,\mathrm{J}.$