Find where e-field is 0 question.

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Homework Statement


Charge 1 is 8nC and is located at x=-1m. Charge 2 is 12nC and is located at x=3m. Find x where e field is equal to 0 due to the 2 charges.


Homework Equations


e=kq/(r^2)


The Attempt at a Solution


The answer is x=1.8, but I just can't seem to get anywhere near that. I know that x has to be between -1m and 3m b/c the E-field due to the 2 charges can only cancel out when x is in between.
E1=e field due to charge 1
E2=e field due to charge 2

0=E1-E2 <<<it's minus because E2 is acting in the negative directive
0=kq1/(x+1)^2 - kq2/(3-x)^2
0=8/(x+1)^2 - 12/(3-x)^2 <<canceled out k and 10^-9 from the nano.
8/(x+1)^2 = 12/(3-x)^2 << the answer is 1.8, I don't want to write out the rest so just plug 1.8 in. When you plug it in it doesn't equal each other.
From that my answer isn't even close to 1.8m. I think I messed up somewhere in the denominator of the equation above, but I can't see how.
 

Answers and Replies

  • #2
Dick
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I don't think you messed up your equation. It looks fine to me. What did you get for an answer? x=1.8m doesn't look like a correct answer to me.
 
  • #3
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I don't think you messed up your equation. It looks fine to me. What did you get for an answer? x=1.8m doesn't look like a correct answer to me.
x=1.8 was the answer in the book and it really makes sense if you draw out the 2 charges.
I'll carry on with my calculations
8/(x+1)^2 = 12/(3-x)^2 << the answer is 1.8, I don't want to write out the rest so just plug 1.8 in. When you plug it in it doesn't equal each other.

8/(x^2+2x+1)=12/(x^2-6x+9)
1/(x^2+2x+1)=1.5/(x^2-6x+9)
now I cross multiplied and got 1.5x^2+3x+1.5=x^2-6x+9
.5x^2+9x-7.5=0
x=.8
 
  • #4
diazona
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...and it really makes sense if you draw out the 2 charges.
What's your reasoning on that? I agree with Dick that x=1.8m doesn't seem like the right answer.
 
  • #5
Dick
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x=1.8 was the answer in the book and it really makes sense if you draw out the 2 charges.
I'll carry on with my calculations
8/(x+1)^2 = 12/(3-x)^2 << the answer is 1.8, I don't want to write out the rest so just plug 1.8 in. When you plug it in it doesn't equal each other.

8/(x^2+2x+1)=12/(x^2-6x+9)
1/(x^2+2x+1)=1.5/(x^2-6x+9)
now I cross multiplied and got 1.5x^2+3x+1.5=x^2-6x+9
.5x^2+9x-7.5=0
x=.8
x around 0.8 is what I get. That puts it at a distance of 1.8 from the charge at x=(-1) and a distance of 2.2 from the charge at x=3. The solution of x=1.8 is a distance of 2.8 from the charge at x=(-1) and a distance of 1.2 from the charge at x=3. That's clearly wrong, right? The balance point should be closer to the weaker charge, yes? Are you sure they didn't ask you how far from the leftmost charge is the balance point?
 
  • #6
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What's your reasoning on that? I agree with Dick that x=1.8m doesn't seem like the right answer.
Sorry my reasoning behind that is poor. I just assumed it was right since it was between -1 and 3m.
x around 0.8 is what I get. That puts it at a distance of 1.8 from the charge at x=(-1) and a distance of 2.2 from the charge at x=3. The solution of x=1.8 is a distance of 2.8 from the charge at x=(-1) and a distance of 1.2 from the charge at x=3. That's clearly wrong, right? The balance point should be closer to the weaker charge, yes? Are you sure they didn't ask you how far from the leftmost charge is the balance point?
The question is exactly as follows "find the point on the x axis where the electric field is 0." Maybe the book made a mistake, it's happened before.
 
  • #7
Dick
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Sorry my reasoning behind that is poor. I just assumed it was right since it was between -1 and 3m.


The question is exactly as follows "find the point on the x axis where the electric field is 0." Maybe the book made a mistake, it's happened before.
It's clear it has to me. It's a distance of 1.8 meters from the left charge. It's not at x=1.8.
 

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