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Physics 21 Online
OpenStudy (anonymous):

Charge of the second particle help please? Force exerted on particle with 5.0x10^-9 C by second particle that is 4 cm away is 8.4x10^-5 N.

OpenStudy (anonymous):

I thought maybe it would be 3.0x10^-9?

OpenStudy (anonymous):

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OpenStudy (anonymous):

do you have the equation for electrical force? or it could be called columbs law?

OpenStudy (anonymous):

OpenStudy (anonymous):

Correct?

OpenStudy (anonymous):

yep! thats the one!

OpenStudy (anonymous):

in the problem we are given: F, q1, r. but we don't know q2 and k. k is a constant, do you have that? and q2 is what we are solving for. so once we get k, we can solve for the answer.

OpenStudy (anonymous):

@J.Velasquez do you have what k equals in your notes?

OpenStudy (anonymous):

I'm not sure if its this one, but I have 8.9 written next to k.

OpenStudy (anonymous):

like this 8.9 x 10^9 N • m2 / C2

OpenStudy (anonymous):

do you have if figured out, or do you should i walk you through the next step?

OpenStudy (anonymous):

@j.velasquez

OpenStudy (anonymous):

I think I can solve, but can you just help me on what are the values? >.< Obviously the K is the 8.9 value but I don't know where the other ones fit in. Thanks for your help btw!

OpenStudy (anonymous):

sorry, back so here are the values we have figured out: F = 8.4x10^-5 k = 8.9 x 10^9 q1 = 5x10^-9 r = 0.04 meters (every time this equation is use, we have to use meters, not cm) q2 = ? [solve for]

OpenStudy (anonymous):

hint, you answer should be between 2.7x10^-9 to 3.5x10^-9

OpenStudy (anonymous):

Thank you so much! (:

OpenStudy (anonymous):

ur welcome

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