Electric field due to two point charge

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Two point charges are placed on the x axis. The first charge, q1 = 8.00nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.0nC , is placed a distance 9.00m from the origin along the negative x axis.

Find the electric field at the origin, point O.
What is Eox, Eoy=?

Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

So here is what i did....
Using Coulombs equation to frind electric field E=kQ/r(sqrd)
I converted from nC to C and found E for both charges.
Then since they're asking for vector component. I plugged the number(E) in to i^hat of both charges, and the j^hat is zero. Then i added it. But its saying that my answer is wrong. can somebody please help me.
What am i missing here? I am probably skipping the most important part of this problem......
 

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  • #2
Doc Al
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Give the x and y components of the electric field as an ordered pair.
They are looking for an answer like: (a, b)
So here is what i did....
Using Coulombs equation to frind electric field E=kQ/r(sqrd)
I converted from nC to C and found E for both charges.
Then since they're asking for vector component. I plugged the number(E) in to i^hat of both charges, and the j^hat is zero. Then i added it.
Show exactly what you did and maybe we can spot the error.
 
  • #3
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I'm working on that EXACT same problem (we're probably working on the same masteringphysics assigment). This is what I got so far.

I start by "placing" a test charge at the origin.

Here's what I think I should do:

E = [tex]\frac{kQ_1}{r^2}[/tex] + [tex]\frac{kQ_2}{r^2}[/tex]

E_1 = [tex]\frac{(9x10^9)(8x10^-9)}{16^2}[/tex] = -0.281 N/C (it's negative because it's giving a force towards the negative x - axis)

E_2 = [tex]\frac{(9x10^9)(6x10^-9)}{9^2}[/tex] = 0.667 N/C

so 0.667 - 0.261 = 0.451.

Probably wrong.
 
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  • #4
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Hey Doc Al did you get this problem figured out. As far as i can tell we're all using the same approach...but in the end no success. Any more ideas or approaches to this problem?
 
  • #5
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Hey Doc Al did you get this problem figured out. As far as i can tell we're all using the same approach...but in the end no success. Any more ideas or approaches to this problem?
did you get the same answer as I did for Part A?
 
  • #6
Doc Al
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Hey Doc Al did you get this problem figured out. As far as i can tell we're all using the same approach...but in the end no success. Any more ideas or approaches to this problem?
The approach spelled out by cse63146 is correct, but the answers are wrong. Two arithmetic mistakes were made: the calculation for E_1 has an error, as does the final subtraction.
 
  • #7
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Found the mistake for E_1; it's 0.281 instead of 0.261

so the final subtraction would be 0.667 - 0.281 = 0.386
 
  • #8
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So in the second case, where q2 is negative (q2 = -6 nC), since it's negative, it would attract the test charge toward the negative x - axis, and thus the electrical field would be negative as well.

So it would be: - 0.667 - 0.281 = - 0.948
 
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  • #9
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Thanks everyone!
 

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