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Two charges and electric field?

  • #1
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Homework Statement


Two charged particles are located on the x axis. The first is a charge +Q at x= -a. The second is an unknown charge located at x= +3a. The net electric field these charges produce at the origin has a magnitude of 2kQ/a^2. Explain how many values are possible for the unknown charge and find the possible values.

Homework Equations


E=k(q/r^2)

The Attempt at a Solution


I got the right answer, I'm just not sure if my thinking is right, because I kind of ignored the way the electric is facing. First of all I assumed q is negative and got 2kQ/a^2=kQ/a^2-k1/9a^2 based on the attraction of q to Q. That turns out to be q=-9Q. Then I did the same thing for when q is positive and got 2Qk/a^2=qk/9a^2 - kQ/a^2, again based on the attraction of q and Q. That turns out to q=27Q, which are the right answer. Is this a valid way of doing it or am I thinking of it wrong?
 

Answers and Replies

  • #2
Bystander
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Is this a valid way of doing it or am I thinking of it wrong?
Is there any particular doubt that stands out in your mind?
 
  • #3
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Is there any particular doubt that stands out in your mind?
My main doubt is that I'm not taking into account the way that the electric field is facing due to the two charges, or would that just be a different way of thinking of it? Only my 4th day of E&M so I'm not too great at it yet :/
 
  • #4
haruspex
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My main doubt is that I'm not taking into account the way that the electric field is facing due to the two charges, or would that just be a different way of thinking of it? Only my 4th day of E&M so I'm not too great at it yet :/
Despite the way you worded your argument, trying both signs for the resultant field is precisely what you did.
2kQ/a^2=kQ/a^2-k1/9a^2
2Qk/a^2=qk/9a^2 - kQ/a^2
Correcting the typo in the first equation:
2kQ/a^2=kQ/a^2-kq/9a^2​
the only difference from the second equation is the sign of the whole right hand side - or, equivalently, of the left hand side.
 
  • #5
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Despite the way you worded your argument, trying both signs for the resultant field is precisely what you did.

Correcting the typo in the first equation:
2kQ/a^2=kQ/a^2-kq/9a^2​
the only difference from the second equation is the sign of the whole right hand side - or, equivalently, of the left hand side.
It's always the simplest parts that mess me up D:
But thank you for the help, makes sense then!
 

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