In the figure particle 1 of charge +4 e is above a floor by distance d 1 = 2.70

Question

In the figure particle 1 of charge +4e is above a floor by distance d1 = 2.70 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?

Rank the situations according to the magnitude of the net electric field at the central point, greatest first. If multiple situations rank equally, use the same rank for each, then exclude the intermediate ranking (i.e. if objects A, B, and C must be ranked, and A and B must both be ranked first, the ranking would be A:1, B:1, C:3). If all situations rank equally, rank each as '1'.

A=

B=

C=

D=

Fourth greatest

Fourth greatest

Explanation / Answer

horizontal force = 9*10^9*+4e*+6e/0.008^2 = 3.375*10^15e^2

= 9.639*10^-23 N

2) A = 4 (actually zero)

B = 1

C = 3

D = 2

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