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Conceptual question using Coulomb

  • Thread starter unrstudent
  • Start date
  • #1
1. 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 Correct answer 46.76N
Magnitude to Left

1.50 µC charge Correct answer 157.325N
Magnitude to Right

-2.00 µC charge correct answer 110.565N
Magnitude to left


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



I have it, but I dont know why it works


This what I have
F13=43.14N
F23=67.425N
F12=89.9N

the 6.00 C made sense Fnet=|-89.9+43.14|
The following were completely different
1.50 C should be Fnet=|-89.9+67.425|
but it not instead, I add the two to get 157.325
this does not make sense because unless 6.00 c was negative, the charges should cancel out like they did in first one


Basically, I got the exact opposite magnitudes and direction, i don't understand
 
Last edited:

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
1.50 C should be Fnet=|-89.9+67.425|
No, both are in the same direction (right) so no minus sign. The positive 6 charge pushes to the right and the negative 2 charge pulls to the right.
 
  • #3
gneill
Mentor
20,793
2,773
You're getting your force directions confused.

Naming the charges from left to right q1, q2, q3, then q2 will be pushed to the right by q1 and pulled to the right by q3. So your sums should be +89.9N + 67.4N = 157.3N. Similarly for q3, where q1 and q2 both PULL on q3.
 
  • #4
Thank you so much,

it was right in front of me.
 

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