Question

A point charge q1=+2.00 nC is held stationary at the origin.A second point charge q2=-4.50 nC moves from the point xi=0.300 m,yi=0.400 to the point xf=0.600 m,yf=0.800 m.How much work is done by the electric force on q2?

Answer 1

Δx = 0,600 - 0,300 = 0,300 м
Δy = 0,800 - 0,400 = 0,400 м
r = √ (0,300² + 0,400²) = 0,500 м
| F | = | k·q₁·(-q₂) / r² | = 9·10⁹·2·10⁻⁹·4,5·10⁻⁹ / 0,500² = 3,24·10⁻⁷ H
A = F·r = 3,24·10⁻⁷·0,500 =  1,62·10⁻⁷ Дж