The electric field due to a point charge

  • Thread starter Jay9313
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  • #1
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http://www.monmsci.net/~fasano/phys2/Chap22_10.pdf
The question is on this PDF File. It's 22.8. I get the logic of it, the electric fields will be 0 only when they both have the same magnitude. The math is shown below the problem, but i don't understand it. Could someone solve this and show me how to do this step by step?

(You can see I'm not cheating, because if I was, I would just copy the answer out of that page. I want to understand it and do well in this class.)
 

Answers and Replies

  • #2
Doc Al
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The way things work here is that you give it a shot and we provide hints and help.

Are you familiar with the field from a single point charge?
 
  • #3
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Yeah, It's E=((k)(Q))/r^2
 
  • #4
tiny-tim
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Welcome to PF!

Hi Jay9313! Welcome to PF! :smile:

Which bit do you not understand?

Do you understand (ignoring the constants) 5q/x2 = 2q/(L-x)2 ?
 
  • #5
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The way I was setting this problem up is
(2)/((4 Pi Episilon zero)(x^2)) = (5)/((4 Pi Epsilon zero)(x+L)^2)
Then I would solve, and I used the quadratic formula. I got
(-10 +_ 6.3L)/6

I have no clue how to solve for L, and I don't even know if my math is right
 
  • #6
Doc Al
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Yeah, It's E=((k)(Q))/r^2
Good.

So, at point (x,0), how would you express the field from each charge? What's the distance to each in terms of the givens?
 
  • #7
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I ignored the constants, that was easy. I got to the point where I got a quadratic. 3x^2 + 10xL + 5L^2

Is the quadratic right? How do you solve for a quadratic with two variables? =(
 
  • #8
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Good.

So, at point (x,0), how would you express the field from each charge? What's the distance to each in terms of the givens?
Well the fields would be equal to each other. The distance would be (x+L)^2 right?
 
  • #9
Doc Al
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How do you solve for a quadratic with two variables? =(
What two variables? L is a constant. You'll express your answer in terms of L.
 
  • #10
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What two variables? L is a constant. You'll express your answer in terms of L.
I got (-10L +_ 6.3L)/6 = 0
How would you solve that? Can you even solve it?
 
  • #11
Doc Al
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Well the fields would be equal to each other. The distance would be (x+L)^2 right?
x is the distance from the point to the origin, where q1 is. So what's the distance to q1?

L is the distance between the charges, so what's the distance to q2?
 
  • #12
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x is the distance from the point to the origin, where q1 is. So what's the distance to q1?

L is the distance between the charges, so what's the distance to q2?
Oh! Is is L and x + L?
 
  • #13
Doc Al
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Oh! Is is L and x + L?
Not quite. The distance to q1 is x. Is the distance to q2 shorter or longer?
 
  • #14
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Oh ok, it's x and x + L , but since it is a function that is squared, it would have to be x-L. So you get (5)/((4 Pi Epsilon zero)(X^2) and you get (2)/(( 4 Pi Epsilon zero)(X-L)^2) right? =D

Then once you cancel terms, you get
5x^2-10xL+5L^2= 2r^2
You then solve the above equation using the quadratic formula, and you get
(10L+2LSqrt(10))/6
The above equation can be simplified by first reducing the 6..
(5L+LSqrt(10))/3)
You can then pull that 3 out to make it look nicer.
(1/3)(5L+LSqrt(10))
Then you can factor out an L
(1/3)(L)(5+Sqrt(10))
And it's not plus or minus Sqrt(10) because it's length, and you can't have a negative length. Am I right? =D
 
  • #15
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Thank you so much =D I sat down for lunch, and I was waiting for the food, and I was just thinking about it non stop. Then it slowly started to fit together. =D
 

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