Question

Three charges +1µC, +3µC and –5µC are kept at the vertices of an equilateral triangle of sides 60 cm. Find the electrostatic potential energy of the system of charges.

Answer 1

Let three charges 1μC , 3μC and -5μC are kept at the vertices A, B and C respectively of an equilateral triangle ABC of sides length 60 cm.
Then, total potential energy = $\bold{\frac{1}{4\pi\epsilon_0r}[q_A.q_B+q_B.q_C+q_C.q_A]}$

Here, r = 60cm = 0.6 m , $q_A$ = 1μC, $q_B$ = 3μC and $q_C$ = -5μC

Now, total potential energy = 9 × 10⁹/0.6 [ 1μC × 3μC + 3μC × -5μC + -5μC × 1μC]
= 1.5 × 10¹⁰ [ 3 × 10⁻¹² - 15 × 10⁻¹²- 5 × 10⁻¹² ] Joule
= 1.5 × 10¹⁰ [ 3 - 15 - 5] × 10⁻¹² Joule
= 1.5 × -17 × 10⁻² Joule
= -25.5 × 10⁻² joule
= -0.255 Joule