Electric field = 0 at what point on x axis?

  • Thread starter connor02
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  • #1
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Homework Statement



you have a +3C and a -1C charge on the x axis. the +3C charge is at the origin and the -1C is at x=1. At what point on the positive x axis does E=0?

Homework Equations



E=kQ/r^2


The Attempt at a Solution



So i have

E=(-1k/r^2)+(3k/(1+r)^2)

Setting E=0

1k/r^2=3k/(1+r)^2

r= sqrt(0.5) = 0.707

is this correct?
 

Answers and Replies

  • #2
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Will your answer be a point in between the two charges, or to the right of each of them?
 
  • #3
ehild
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The Attempt at a Solution



So i have

E=(-1k/r^2)+(3k/(1+r)^2)

What does "r" mean?

Setting E=0

1k/r^2=3k/(1+r)^2

r= sqrt(0.5) = 0.707

is this correct?

No, r= sqrt(0.5) is not correct. How did you get it?

ehild
 
  • #4
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ah, r=1.707?
 
  • #5
ehild
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What is r?
 
  • #6
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r=the point on the x axis where the E firld is 0

right?
 
  • #7
ehild
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You used the expression of the electric field at distance "r" from a point charge. "r" is distance, not a point. What do you substitute for "r" for the 3C and -1C charges?


ehild
 
  • #8
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my bad. r=distance to the right of the -1C charge. x=1.707 is the point on the x axis where E=0.

right?
 
  • #9
HallsofIvy
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r=the point on the x axis where the E firld is 0

right?
No, it isn't. Your r is clearly a number, not a point. I suspect you mean it is the x-coordinate of such a point. In that case, the distance to 0 is r but the distance to 1 is |1- r| not 1+ r.
 
  • #10
ehild
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my bad. r=distance to the right of the -1C charge. x=1.707 is the point on the x axis where E=0.

right?

If so, your equation 1k/r^2=3k/(1+r)^2 is correct, but the solution is wrong. Check it.

ehild
 
  • #11
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WOW i feel STUPID.

r=-0.366 or 1.366

Since x is to the right of the -1C charge, x = 2.366 when E=0.

I hope i am right now.
 
  • #12
ehild
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It is correct now. Well done!

ehild
 
  • #13
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Thanks man!

Now, I'm just curious, but if I wanted to find the spot where V=0, I would have

0=3Q/(1+r) + (-Q/r)

so r=1/2

Does that mean V=0 at x=1.5 or V=0 at both x=1.5 and x=0.5?
I think it is the second one as all potential lines have to form a closed surface.
Am I right?

thank you.
 
  • #14
ehild
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It is less confusing if you calculate with the position of the point on the x axis, that is with its x coordinate.
If a point charge Q is at position x=a, its contribution to the potential at x is kQ/|x-a|. Note the absolute value in the denominator!

In this problem, U(x)=3k/|x|-k/|x-1|. U(x)=0 at x=1.5 and x=3/4.

ehild
 
  • #15
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Yes you are right, it is a lot easier to make sense of the answer if i use the point on the x axis.

Why did you use absolute value?

I used

0 = 3k/x + [(-Q)/(x-1)]

and i get the solution x = 1.5 only

thanks.
 
  • #16
ehild
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The term "r" in the denominator of the potential is distance. The distance is positive (or zero). The coordinate can be both positive and negative.
At distance r from a point charge, the potential is the same either at the right hand side of the charge or at the left hand side.
Find the potential of the -1 C charge at x=3/4. What do you get without absolute value and what is the right value of the potential?

ehild
 

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