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Magnitude (electric charge: need correction)

  • Thread starter BunDa4Th
  • Start date
  • #1
188
0

Homework Statement


Calculate the magnitude and direction of the Coulomb force on each of the three charges shown in Figure P15.10.

http://www.webassign.net/sf5/p15_10.gif

thats 3 cm if its to small to see.

6.00 µC charge
Magnitude

1.50 µC charge
Magnitude

-2.00 µC charge
Magnitude

Homework Equations



F = K|q1||q2|/r^2

The Attempt at a Solution



I tried this problem and it said i was incorrect but within 10% to the answer and that my rounding is off or something.

this is what i did:

8.9875 x 10^9 (6 x 10^-6)(2 x 10^-6)/.05^2
(converted to m and C)

.10785/.05^2

= 43.14 N

That looks correct but it is saying im off. It say that im within 10% to the answer. Can anyone look at this and tell me what im doing incorrectly?

For the other 2 I am not sure what to do, but i gave it a try

for 1.5uC i did

K(1.5 x 10^-6)(2 x 10^-6)/.02^2 this completely got me the wrong answer and not sure where to start on this.
 

Answers and Replies

  • #2
hage567
Homework Helper
1,509
2
There will also be a force from q3 acting on q1. You must sum the forces from each one.
 
  • #3
188
0
okay when you say q3 acting on q1 do i do the same thing i did up top? that i did for 6uC because i dont think i quiet understand it.

also from the order left to right how would the order of q1, q2, and q3 be?
 
  • #4
188
0
Okay, I understand what you meant by force from q3 acting on q1 now is. I was able to figure this out now. Thanks for that help that i would probably never would have thought of.

I still have trouble figuring out how to get -2uC. How would i go about solving this? I understand how to get the other two but this one seem to be different or would it be the sum of the two forces i found?
 
  • #5
hage567
Homework Helper
1,509
2
You would use the same method as the others. Remember that Coulomb's law obeys Newton's third law, so the force exerted on q1 by q2 is equal and opposite to the force of q2 on q1. Keep in mind that force is a vector quantity. Since one of the charges is negative you have to figure out if the force between the charges you are using is attractive or repulsive.
 
  • #6
188
0
Thanks for the explanation on that and i was able to get it and it was attractive charges because of the different sign.
 

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