I would like to build a two particle system I begin by bring

I would like to build a two particle system. I begin by bringing in particle A from infinity. Particle A has a charge of 6.3 Times 10^-6 C. I then go back out to infinity and bring in particle B. Particle B has a charge of -3.1 Times 10^-8 C. If I place particle B 3.7cm away from particle A, what is the potential energy of this two-particle system?

Solution

Charge of A is q = 6.3 x10 ^-6 C

Charge of B is q \' = -3.1x10 ^-8 C

Distance between two charges A and B is r = 3.7 cm = 3.7 x10 ^-2 m

The potential energy of two particle system is U = Kqq\'/r

where K = Coulomb\'s constant = 8.99 x10 ⁹ Nm ²/C ²

Substitute values you get U = (8.99x10 ⁹) (6.3 x10 ^-6 )(-3.1 x10 ^-8 ) /(3.7 x10 ^-2)

                                         = -0.04745 J

                                         = 47.45 x10 ^-3 J