In the figure below, particle 1 of charge +4e is above a floor by distance d1 =
ID: 1916089 • Letter: I
Question
In the figure below, particle 1 of charge +4e is above a floor by distance d1 = 1.60 mm and particle 2 of charge +6e is on the floor, at distance d2 = 6.00 mm horizontally from particle 1. What is the x component of the electrostatic force on particle 2 due to particle 1?
In the figure below, particle 1 of charge +4e is above a floor by distance d1 = 1.60 mm and particle 2 of charge +6e is on the floor, at distance d2 = 6.00 mm horizontally from particle 1. What is the x component of the electrostatic force on particle 2 due to particle 1?Explanation / Answer
X-component of the electrostatic force on particle 2 due toparticle 1 is F = Kq2q1 / (d2) ^ 2 Where K = 8.99 * 10 ^ 9 N m^ 2/ C ^ 2 q1= 4e = 4 * 1.6 * 10 ^ -19 C q2= 6e = 6 * 1.6 * 10 ^ -19 C d2 = 6 mm = 6 * 10 ^ -3 m plug the values we get F = 1.534 * 10 ^ -22N