Question

Three equal 1.75-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.250 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.

Answer 1

`U = 0.33 J`

Step-by-step explanation:

As we know that electrostatic potential energy of a system of charge is given by equation

`U = (kq_1q_2)/(r)`

here we know that

`q_1 = q_2 = 1.75 \mu C`

now we also know that

r = distance between two charges = 0.250 m

now here three charges are placed at the vertices of the triangle

so here net electric potential energy of all three charges is 3 times the energy of one pair of charge

`U = (3kq_1q_2)/(r)`

`U = (3(9* 10^9)(1.75 \mu C)^2)/(0.250)`

`U = 0.33 J`