Question

A point charge q₁ = +3.00 µc is held stationary at the origin. a second point charge q₂ = -4.30 µc moves from the point x = 0.110 m, y = 0 to the point x = 0.250 m, y = 0.250 m. how much work is done by the electric force on q₂?

Answer 1

the work done by the electric force on the second charge is W= 0.72 J

Step-by-step explanation:

the initial distance d₁ between the charges is

d₀=√(x₀² + y₀²)= √[(0.110m)² + 0²] = 0.110 m

the final distance d between the charges is

d=√(x² + y²)= √[(0.250m)² + (0.250m)²] = 0.353m

the work done by the electric force on the second charge is

W= k*q₂*q₁*(1/d-1/d₀)

where

W= work that was done by the electric force on the second charge

k = coulomb's constant = 8.987*10⁹ N·m²/C²

q₂ and q₁ = electric charge of the second and first charge respectively

replacing values

W= k*q₂*q₁*(1/d-1/d₀) = 8.987*10⁹ N·m²/C²*(-4.30*10⁻⁶ C)*(3.00*10⁻⁶C)*( 1/0.353m - 1/0.110 m) = 0.72 N*m = 0.72 J

W= 0.72 J