Question

Three equal point charges, each with charge 1.25 μC , are placed at the vertices of an equilateral triangle whose sides are of length 0.450 m . What is the electric potential energy U of the system

Answer 1

Answer:

electric potential energy  is 9.36 ×$10^{-2}$ J

Explanation:

given data

charge q = 1.25 μC =1.25×$10^{-6}$ C

length L = 0.450 m

to find out

electric potential energy

solution

we apply electric potential energy U formula

U = U1 + U2 + U3   .................1

we know magnitude of all three charge is

q = q1 = q2 = q3

so here from 1

U = 1 / 4π∈ × (q²/L + q²/L +q²/L )

U = 1 / 4π∈ × 3(q²/L)

put here all value and we know ∈ = 8.85 ×$10^{-12}$ F/m

U = 1 / 4π( 8.85 ×$10^{-12}$) × ( 3 (1.25×$10^{-6}$)² / 0.450 )

U = 8.99×$10^{9}$ × ( 3 (1.25×$10^{-6}$)² / 0.450 )

U = 9.36 ×$10^{-2}$

so electric potential energy  is 9.36 ×$10^{-2}$ J