In the figure particle 1 of charge +6e is above a floor by distance d1 = 2.10 mm

Question

In the figure particle 1 of charge +6e is above a floor by distance d1 = 2.10 mm and particle 2 of charge +8e is on the floor, at distance d2 = 7.50 mm horizontally from particle 1. What is the x component of the electrostatic force on particle 2 due to particle 1?

In the figure particle 1 of charge +6e is above a floor by distance d1 = 2.10 mm and particle 2 of charge +8e is on the floor, at distance d2 = 7.50 mm horizontally from particle 1. What is the x component of the electrostatic force on particle 2 due to particle 1?

Explanation / Answer

F = k*4e*8e/R^2
= k*32*e^2/(d1^2+d2^2) =9*10^9*48*(1.6*10^-19)^2/(2.1^2+7.5^2)/10^-6

=1.823*10^-22 N

tan(alpha) = d1/d2 =2.1/7.50

alpha = 15.64 degrees

Fx= F*cos(alpha) = 1.823*10^-22*cos(15.642) =1.755*10^-22 N

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