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

Answer:

0.12959085 J

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

$U=k\dfrac{q^2}{d}$

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

$U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}$

$\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J$

The potential energy of the system is 0.12959085 J