Question

Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 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

0.12959085 J

Step-by-step explanation:

k = Coulomb constant =
`8.99* 10^(9)\ Nm^2/C^2`

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

`U=k(q^2)/(d)`

In this system with three charges which are equidistant from each other

`U=k(q^2)/(d)+k(q^2)/(d)+k(q^2)/(d)`

`\\\Rightarrow U=k(3q^2)/(d)\\\Rightarrow U=8.99* 10^9* (3* (1.55* 10^(-6))^2)/(0.5)\\\Rightarrow U=0.12959085\ J`

The potential energy of the system is 0.12959085 J