Electrostatic force-Coulombs law

AI Thread Summary
A user is attempting to solve a problem involving electrostatic forces between multiple charges using Coulomb's law. They calculated the forces acting on a -12 µC charge due to a 64 µC charge and a 16 µC charge, arriving at magnitudes of 74.72 N and 691.2 N, respectively. After breaking these forces into components and summing them, they obtained a resultant force of 707.32 N, which contradicts the textbook answer of 2650 N. Other participants suggest that the user's method is correct and question the accuracy of the textbook. The discussion highlights the potential for errors in textbook solutions and the importance of verifying calculations in physics problems.
Thrax
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Homework Statement



Hi, I am working on the following question.
A 64 µC charge is located 30cm to the left of a 16 µC charge. What is the resultant force on a -12 µC charge positioned exactly 50mm below the 16 µC charge?




Homework Equations


Coulombs law



The Attempt at a Solution


I see a right triangle So I tried to solve using the vector components as follows
Fac=9.0e9(65x10e-6 * 12x10e-6)/(.3^2+.05^2) = 74.72C
Fbc = 9.0e9(16x10e-6 * 12x10e-6)/(.05^2) = 691.2C


For the rest, I'm not sure if I'm attacking this thing properly to solve for the vectors? I've tried at least 6 times... but no luck.

I could use a little direction...I attached a picture of the problem. My be helpful if it show's up.
 

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You have calculated the magnitudes of the vectors. Now you need to add them as vectors to find the resultant force. If you tried this without success, you need to show what you did so we can point out where you went wrong. By the way, force is measured in Newtons, abbreviation N, not C.
 
OK...heres what I've done. Flipped the triangle a little setting C and B on the x axis. I'm new to all this stuff so please forgive my ignorance.
Fbc = 691.2
Fac = 74.72

Fbcx = 691.2
Fbcy = 0

Facx = 74.72N cos(80) = 12.28
Facy = 74.72 sin(80) = 73.58

Fbcx + Fbcy
691.2 + 12.28 = 703.48
Fbcy + Facy
0 + 12.28 = 73.58
Finally
Fc = Sqrt(703.48 ^2 + 73.58^2) = 707.32N
Not sure where I’m going wrong…but I know I’m incorrect.
 
I didn't check your arithmetic, but your basic method looks OK. How do you know it's incorrect?
 
The textbook I have gave an answer. According to them the correct answer is 2650N, 113.3deg.
 
I get 707.3 for the resultant. Looks like you are OK here.
 
Thrax said:
The textbook I have gave an answer. According to them the correct answer is 2650N, 113.3deg.
The textbook answer is way off. (What textbook?)
 
Thrax said:
The textbook I have gave an answer. According to them the correct answer is 2650N, 113.3deg.
Unless both you and I made the same mistake exactly, the problem as stated has the answer that you gave. Are you sure you posted the problem as it appears in the book? If yes, then you need to consider that sometimes the back of the book has the wrong answers. In physics if you're right, you're right.
 
The textbook is called. "Physics" 7th edition by Paul E. Tippens. Mc Graw Hill
 
  • #10
Yes, I typed it exactly as it appears.
 
  • #11
You spend so much on textbooks...You'd think the least they could do is edit!
 
  • #12
If you guys think it’s OK...than its OK with me! Thanks for taking a look!
 
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