Coulomb's Law - Four Charges with Coordinates are Given, Find the Electrostatic Force


by antiderivativ
Tags: charges, coordinates, coulomb, electrostatic, force
antiderivativ
antiderivativ is offline
#1
Aug4-09, 10:27 PM
P: 17
Four point-charges are fixed at the corners of a 3.0m X 4.0m rectangle. The coordinates of the corners and the values of the charges are listed below.
q1 = 100 microC (0, 4m), q2 = 36 microC (4m, 3m), q3 = 125 microC (0, 3m) and q4 = 32 microC (0,0). Compute the net electrostatic force acting on the 100 microC charge.
ke = 8.99 x 109

I'm using Coulomb's Law.
F = [tex]\frac{ke*q1*q2}{r^2}[/tex]

Here is my attempt at a solution. Is it correct?

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alphysicist
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#2
Aug5-09, 10:58 PM
HW Helper
P: 2,250
Hi antiderivativ,

Quote Quote by antiderivativ View Post
Four point-charges are fixed at the corners of a 3.0m X 4.0m rectangle. The coordinates of the corners and the values of the charges are listed below.
q1 = 100 microC (0, 4m), q2 = 36 microC (4m, 3m), q3 = 125 microC (0, 3m) and q4 = 32 microC (0,0). Compute the net electrostatic force acting on the 100 microC charge.
ke = 8.99 x 109

I'm using Coulomb's Law.
F = [tex]\frac{ke*q1*q2}{r^2}[/tex]

Here is my attempt at a solution. Is it correct?

No, I don't believe that is correct. When you find the components of F13, you are using an angle of 45 degrees. It would have been a 45 degree angle if the charges were at the corners of a square, but since this is a rectangle it will be different.

You can use your 3-4-5 triangle you have on the page to find the correct angle. What do you get?
antiderivativ
antiderivativ is offline
#3
Aug6-09, 12:11 AM
P: 17
Thanks for the reply! My new angle is 53.13. Is this better? :)

alphysicist
alphysicist is offline
#4
Aug6-09, 07:52 AM
HW Helper
P: 2,250

Coulomb's Law - Four Charges with Coordinates are Given, Find the Electrostatic Force


Quote Quote by antiderivativ View Post
Thanks for the reply! My new angle is 53.13. Is this better? :)
Really close! But you changed your triangle when you calculated the angle, and that gave you the wrong angle.

If you look at the 3-4-5 triangle about halfway down the page on the left side, the 3 side is vertical and the 4-side is horizontal, and that matches your problem and calculation.

At the bottom of the page, you switched the 3 and 4 sides. You did the correct procedure; it's just that if you use your original triangle, you'll do:

[tex]
\tan^{-1}\left(\frac{3}{4}\right)
[/tex]

instead of the arctangent of 4/3.


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