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Electric Field

  • Thread starter eestep
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  • #1
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Homework Statement


A charge of -11.8 micro-Coulombs is placed on x axis at x=0 meters. Another charge of -7.3 micro-Coulombs is placed on x axis at x=+.42 meters. Other than infinity, where on x axis is total electric field equal to zero? Answer in centimeters.
q1=-11.8[tex]\mu[/tex]C
q2=-7.3[tex]\mu[/tex]C

Homework Equations


E=kq/r2[tex]\widehat{}r[/tex]


The Attempt at a Solution


E1=E1[tex]\widehat{}x[/tex]=-(8.99*1099Nm2/C2)(11.8*10-6C)/r12[tex]\widehat{}x[/tex]
E2=E2[tex]\widehat{}x[/tex]=(8.99*109Nm2/C2)(7.3*10-6C)/r22
Ex=E1x+E2x=-(8.99*109Nm2/C2)(11.8*10-6C)/r12+(8.99*109Nm2/C2)(7.3*10-6C)/r22=0
11.8/7.3=r12/r2^2
r1/r2=[tex]\sqrt{}11.8/7.3[/tex]
 

Answers and Replies

  • #2
31
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I didn't check all of your steps. But if that is right then what you want now is to just have the x-coordinate. So really you only want one 'r', the other one can be written in terms of this r as say, r2 = 0.42 - r1.
Then you can just solve for your r which is the x-coordinate of the equilibrium point.
 
  • #3
36
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How do I do that? I get r2^2=7.3r1^2/11.8.
 
  • #4
31
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Well if the ratio is correct:
[tex]\frac{r_{1}}{r_{2}} = \frac{\sqrt{11.8}}{7.3}[/tex]
Just look at the problem in question. Your first charge is at x=0 the second charge is at x=0.42
Now you are asked to find the equilibrium point that should just be a single coordinate. So from the origin to the equilibrium point is r let's say, this also happens to be the distance from the first charge as well since the first charge happens to be at the origin which you called [itex]r_{1}[/itex]. So you have a relation between the distance from the first charge to the equilibrium point and the coordinate of the equilibrium point... which is just: [itex]r_{1} = r[/itex].
Now you want to express the distance from the second charge to the equilibrium point. It should be obvious that it is: [itex]r_{2} = 0.42 - r[/itex] which is the same as: [itex]r_{2} = 0.42 - r_{1}[/itex]

If it's not obvious then draw a picture. Put the equilibrium point between the two charges and come up with those expressions for the distances to each charge, where r is the unknown x coordinate of the equilibrium point.

So therefore your ratio becomes:
[tex]\frac{r}{0.42 - r} = \frac{\sqrt{11.8}}{7.3}[/tex]
Then you just do some algebra and solve for r

you should get r = <some number> that does not depend on [itex]r_{1}[/itex] or [itex]r_{2}[/itex].
 
  • #5
36
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I appreciate your aid!
 

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