Question

Can anyone solve this for me please

Spamers please stay away ​

Answer 1

Potential energy = [(k ∙ Q1Q2)/r] here q = 10μc at corners of equilateral triangle r = 10 cm = 0.1m ∴ Potential energy for pals of charge = [{k(q ∙ q)}/r] + [{k(q ∙ q)}/r] + [{k(q ∙ q)}/r] = [(3k)/r]q2 = 3 × [1/(4π∈0)] × (1/0.1) × (10 × 10–6)2 = 3 × 9 × 109 × 10 × 10–10 = 27J