Three charges are placed at the vertices of an equilateral triangle of side a =

Question

Three charges are placed at the vertices of an equilateral triangle of side a = 0.71 m, as shown in the figure below. Charges 1 and 3 are +5.1

Three charges are placed at the vertices of an equilateral triangle of side a = 0.71 m, as shown in the figure below. Charges 1 and 3 are +5.1 mu C; charge 2 is -5.1 mu C. Do you expect the net force acting on charge 1 to have a magnitude greater than, less than, or the same as the magnitude of the net force acting on charge 2? Find the magnitude of the net force acting on charge 1. Find the magnitude of the net force acting on charge 2.

Explanation / Answer

Part A)

They will all be the same. Same charges, same distances of separation, and same parts will cancel and same parts will add.

Part B and C

F = kqq/r^2

F = (9 X 10^9)(5.1 X 10^-6^2/(.71)^2

F = .464 N

The adding component = .464(sin 30) = .232

Net force = 2(.232) = .464 N

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