I have worked this out several times and keep ending up with either 4.42 or 44.2

Question

I have worked this out several times and keep ending up with either 4.42 or 44.2. I would appreciate a detailed step by step instruction of how to calculate this.

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.932 g, q = 2.76 C is located on the x axis at x = 22.0 cm, moving with a speed of 73.1 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.) Number T44.2

Explanation / Answer

m = 0.932 g, q = 2.76 µC, x = 22.0 cm, v = 73.1 m/s, find Q

F = kqQ/x2 = mv2/x

Q = mv2x/(kq) = [0.932*(73.1)^2*22*0.01] / 9*10^9 *2.76 = 4.900 C

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