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

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