Four charges are placed at the corners of a square. The side of the square is a

Question

Four charges are placed at the corners of a square. The side of the square is a = 0.32 m.

a) If the charge are identical and positive q = 44.0 x 10-9 C what would be the magnitude of the force acting on each of the charges? Answer in Newtons.

b) We change the charges by q1 = 44.0 x 10-9 C, q2 = -24.0 x 10-9 C, q3 = 26.0 x 10-9 C, q4 = -46.0 x 10-9 C. What would be the magnitude of the force acting on charge q3? Answer in Newtons.

c) We place the same charges at the corners of a rectangle of sides a and b (instead of a square). If b=2a/3 what would be the magnitude of the force acting on charge q3? Answer in units of Newtons.

Explanation / Answer

since two are at a distance of .32 and one at the diagonal at the distance of .32*sqrt(2)

force on each charge = 2kQ^2 / r1^2 + kQ^2/r2^2 = 2*9*10^9*(44*10^-9)^2/ .32^2 + 9*10^9*(44*10^-9)^2 / (.32*sqrt(2) )^2 = 3.4*10^-4 + .85 * 10^-4 = 4.25*10^-4 N

F = 9*10^-9 (- 44* 24)/.32^2 - 9*10^-9(44*46)/.32^2 + 9*10^-9(44*26)/2*.32^2 = -2.2*10^-4 N

Now the above equation can be used by just modifying the distances

F = 9*10^-9 (- 44* 24)/.32^2 - 9*10^-9(44*46)/(2/3*.32)^2 + 9*10^-9(44*26)/(sqrt(13/9)*.32)^2 = (-92812.5 - 400253.9 + 5354.5) *10^-9 = -4.877 * 10^-4 N

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