In the figure particle 1 of charge +5 e is above a floor by distance d 1 = 3.00

Question

In the figure particle 1 of charge +5e is above a floor by distance d1 = 3.00 mm and particle 2 of charge +6e is on the floor, at distance d2 = 8.00 mm horizontally from particle 1. What is the xcomponent of the electrostatic force on particle 2 due to particle 1?

In the figure particle 1 of charge +5e is above a floor by distance d 3.00 mm and particle 2 of charge +6e is on the floor, at distance d2 8.00 mm horizontally from particle 1. What is the x component of the electrostatic force on particle 2 due to particle 1? Number Units

Explanation / Answer

d21 = sqrt(d2^2 + d1^2) = sqrt(8^2 + 3^2)mm = 8.544mm = 0.00854m

F21 = k*q2*q1/d21^2 = 8.99*10^9*5*6*(1.6*10^-19)^2 / 0.00854^2

F21 = 9.47*10^-23 N

F21 is at angle arctan(3/8) below the horizontal = 20.556 degrees

So x component of F21 is 9.47*cos(20.556) * 10^-23 N = 8.87*10^-23 N

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