Electric Charges Causing Centripetal Motion

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mdf730
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Homework Statement



A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.671 g, q = 5.90 µC is located on the x-axis at x = 15.6 cm, moving with a speed of 54.4 m/s in the positive y direction. For what value of Q (in μC) will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

Homework Equations



F=F
F=kqQ/r^2
F=mv^2/r

The Attempt at a Solution



F=F
kqQ/r^2=m(v^2)/r
kqQ=m(v^2)r
(8.99*10^9)(5.9*10^-6 C)(Q) = (.000671 kg)((54.4 m/s)^2)(.156 m)
53100 * Q = .30977
Q= .0000058337 C
Q= 5.8 μC

I have not been able to find my error, but I did something wrong, as the wileyplus marks it incorrect.
 
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mdf730 said:

Homework Statement



A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.671 g, q = 5.90 µC is located on the x-axis at x = 15.6 cm, moving with a speed of 54.4 m/s in the positive y direction. For what value of Q (in μC) will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

Homework Equations



F=F
F=kqQ/r^2
F=mv^2/r

The Attempt at a Solution



F=F
kqQ/r^2=m(v^2)/r
kqQ=m(v^2)r
(8.99*10^9)(5.9*10^-6 C)(Q) = (.000671 kg)((54.4 m/s)^2)(.156 m)
53100 * Q = .30977
Q= .0000058337 C
Q= 5.8 μC

I have not been able to find my error, but I did something wrong, as the wileyplus marks it incorrect.
Welcome to PF!

What is the direction of the force? (Hint: attractive or repulsive?).

AM