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 = 0.81 m, as shown in the figure below. Charges 1 and 3 are +6.1

Three charges are placed at the vertices of an equilateral triangle of side a = 0.81 m, as shown in the figure below. Charges 1 and 3 are +6.1 mu C; charge 2 is -6.1 mu C. Find the magnitude and direction of the net force acting on charge 3. If charge 3 is moved to the origin, will the net force acting on it there be greater than, less than, or equal to the net force found in part (a)? Find the net force on charge 3 when it is at the origin.

Explanation / Answer

Part A)

The horizontal components of the charges will cancel, buth the vertical components will add.

F= kqq/r^2

F = (9 X 10^9)(6.1 X 10^-6^2)/(.81)^2

F = .510

The y component = .510(sin 30) = .255

Total F = 2(.255) = .510 N straight down

Part B)

If charge three is moved to the origin the net force will increase since the distance between the charges decreases.

Part C)

F= kqq/r^2

F = 2(9 X 10^9)(6.1 X 10^-6^2)/(.405)^2

F = 4.08 N

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