Electric fields and electrostatic forces

  • #1
Electric fields and electrostatic forces (1 question)

I was wondering if someone can show me how to answer this problem. Thank you.


1)Two charges are placed on the x-axis, an unknown positive charge at x=0 cm and a negative 8.642e-6 C charge at 2.993 cm from the origin. Calculate the magnitude of the charge at x=0 cm which causes the total electric field at 11.917 cm to be zero.
 
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Answers and Replies

  • #2
Tide
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Show us what you did then I'm sure you'll get plenty of help!
 
  • #3
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Electric field,[tex] E = \frac{kq}{r^2}[/tex]
Since you have two particles with charges, lets call the unknown charge q1, and the known charge q2, where

Since E = 0, [tex] \frac{kq_1}{r^2} + \frac{kq_2}{r^2} = 0[/tex], where r is the distance from the charge to the point of zero electric field. Therefore...
For the first charge its .11917m away, and for the second charge its .11917m - .02993m = .08924m.

[tex]\frac{kq_1}{.11917^2} + \frac{k*-8.642e-6}{.08924^2} = 0[/tex]

The k's cancel, and therefore q1 = 1.54e-5 C.

I'm not 100% sure if I did this right, so wait for more responses before taking my word for it. :smile:
 
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  • #4
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Conceptually it is correct although I haven't done the calculation so...
 

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