Finding the Charge for Circular Motion in an Electric Field

[nothingatall]

Homework Statement

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.776 g, q = 5.04 µC is located on the x-axis at x = 18.7 cm, moving with a speed of 39.8 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=mv^2/r- centripetal force
F=k*q1*Q/r^2

The Attempt at a Solution

i'm thinking of starting by setting the two forces together and find the other Q. Just making sure if its the correct way to find it. If its not can someone outline the steps i need to follow? thanks.

 

[tiny-tim]

Hi nothingatall!

(try using the X² tag just above the Reply box )

Yes, that's exactly the way to do it.

 

[nothingatall]

ok so i tried it and apparently my answer is wrong. Here's my work:

Fc= (7.76e-4kg)(39.8m/s)^2/18.7e-2m
6.573=(8.99e9)(5.04e-6C)(Q)/(18.7e-2m)^2
=5.07uC.

it asks for the answer in microC so I don't know where i messed up. I'm at my last attempt before it get marked wrong and my professor is no help.

 

[tiny-tim]

ok so i tried it and apparently my answer is wrong. Here's my work:

Fc= (7.76e-4kg)(39.8m/s)^2/18.7e-2m
6.573=(8.99e9)(5.04e-6C)(Q)/(18.7e-2m)^2
=5.07uC.

it asks for the answer in microC so I don't know where i messed up. I'm at my last attempt before it get marked wrong and my professor is no help.

hmm … using your figures of 7.76*(.398)²*0.187/8990*5.04, I get 5.07 10^-6 also.

Does the sign matter?

 

[nothingatall]

I'm not sure but I'm afraid to try it on account of its my final attempt :(

 

[ideasrule]

I'll take a gamble and say that yes, the sign does matter, so the answer is -5.07 10^-6 C.