Finding the Net Electric Force on a Charge

  • Thread starter Tekmachine
  • Start date
  • #1

Homework Statement


Three positive point charges are arranged in a triangular pattern in a plane, as shown below.
The Coulomb constant is [tex]8.98755*10^9 N · m^2/C^2.[/tex]

http://img150.imageshack.us/img150/1052/electricforceproblemqp7.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

Find the magnitude of the net electric force on the 6 nC charge. Answer in units of N.

Homework Equations



[tex]F = \frac{K(qQ)} {r^2}}[/tex]

The Attempt at a Solution



Pythagorean Theorem: [tex]\sqrt{(6)^2 + (6)^2} = 8.485[/tex]

[tex]\frac{(8.98755*10^9 N · m^2/C^2)(6*10^{-9}C)(2*10^{-9}C)} {(8.485m)^2}} = 1.498*10^{-9} N[/tex]

[tex]\frac{(8.98755*10^9 N · m^2/C^2)(6*10^{-9}C)(9*10^{-9}C)} {(8.485m)^2}} = 6.7411*10^{-9} N[/tex]

[tex]sin(45)(1.498*10^{-9} N + 6.7411*10^{-9} N) = 5.8259*10^{-9} N[/tex]

Unfortunately that's not the right answer :frown:
 
Last edited by a moderator:

Answers and Replies

  • #2
532
2
You made a good start so far. The magnitude of the force on the 6 nC particle due to the other two particles is done correctly. Your next step is wrong.

Do the following: Draw the two force vectors on the 6 nC charge, and then determine their x- and y- components. Now add the components together to get the net force components on the 6 nC charge. Apply the Pythagorean theorem to these components to get the magnitude of the net force.
 
  • #3
Attempt #142,032:
[tex]F1 = cos(45)1.498*10^{-9} N[/tex] x [tex] + sin(45)1.498*10^{-9} N[/tex] y
[tex]F2 = -cos(45)6.741*10^{-9} N[/tex] x [tex] + sin(45)6.741*10^{-9} N[/tex] y
[tex]Fnet = 3.706*10^{-9} N[/tex] x [tex] + 5.825*10^{-9} N[/tex] y

[tex]||Fnet|| = 6.905*10^{-9} N[/tex]

Thanks Guys!
 
  • #4
1
0
One of the y components should be negative, not one of the x components. The magnitude still comes out the same.
 

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