What is the charge in the middle of the square?

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Homework Statement



As shown in the figure, a square has sides of d = 9.0 cm with a charge of Q = +18.0 µC at one corner (a) and with charges of q = -6.0 µC at the remaining three corners (b, c, and d).

Image: http://www.webassign.net/grrphys2/17-p-018-alt.gif" [Broken]

Find the electric field at the center of the square.
magnitude _____ N/C

Homework Equations



E=kq/r^2

The Attempt at a Solution



Since the charges at corners b and d are the same, they cancel each other out, they won't contribute to the field in the middle. So the only corners we have to worry about are corners a and c. Ea = 18 uC/(4 pi eps d^2)
From the charge in c, the field strength is Ec = -6 uC/(4 pi eps d^2)

So the total field (both are directed toward c) = (12 uC) / (4 pi eps0 d^2).

The only thing I don't know how to calculate is r. I tried 4.5 sqrt. 2. I tried 0.09. I tried 3 sqrt. 2. None of them worked. Can you please help me?

I only have one more answer to try. I just thought of something, what if I try 4 sqrt (0.09 m)? Will that work?
 
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Answers and Replies

  • #2
Doc Al
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Since the charges at corners b and d are the same, they cancel each other out, they won't contribute to the field in the middle. So the only corners we have to worry about are corners a and c.
Good!
The only thing I don't know how to calculate is r. I tried 4.5 sqrt. 2. I tried 0.09. I tried 3 sqrt. 2. None of them worked. Can you please help me?
Careful with units. Express the length in meters, not cm. (The first answer is correct, but it's in cm.)
 
  • #3
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Oh, so it should be 0.045 m sqrt. 2? Thank you so much, I wasn't paying attention to my units. This forum is awesome, I get such quick helpful replies, and I learn, not just copy.
 
  • #4
Doc Al
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Oh, so it should be 0.045 m sqrt. 2?
Right!
 
  • #5
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Well, it says it's wrong. And that was my last submission :frown:

Thanks anyway for the quick reply.
 
  • #6
Doc Al
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Well, it says it's wrong. And that was my last submission
Oh well. But at least you had the distance right!

ah... I see the problem. For some reason you subtracted the field from each charge. But those fields act in the same direction, so they add up.

(My apologies for not looking at the rest of your post that carefully. :uhh:)
 
  • #7
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I did add them. 18 + -6 = 12.
I still don't understand what I did wrong.
 
  • #8
Doc Al
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I did add them. 18 + -6 = 12.
I still don't understand what I did wrong.
The problem is that you cannot blindly plug the charge into the formula E = Q/(4πε r2) to get the field, since the direction of the field depends on where you are with respect to the charge.

My recommendation is to use the formula to get the magnitude of the field, then use the sign of the charge to get the direction of the field. In this case:
- The charge at a is positive, so the field points away from a. Thus, at the midpoint, the field from charge Qa points toward c.
- The charge at c is negative, so the field points toward c. Thus, at the midpoint, the field from charge Qc points toward c.

To get the total field at the midpoint, you'll need to add the magnitudes, since they are pointing in the same direction.

Make sense?
 
  • #9
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Yes, the third part of the question was in what direction, and I already answered in "towards corner C" and they said it was correct.
I understand everything you said, it makes sense to me, but that's not enough to pass the class, I have to get the number right also. I have no way of checking my answer now because I used up all my submissions. I've been adding the fields from submission 1, and it says it's incorrect.
 
  • #10
Doc Al
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I've been adding the fields from submission 1, and it says it's incorrect.
What you added was what you got from plugging into the formula:
Ea = 18 uC/(4 pi eps d^2)
From the charge in c, the field strength is Ec = -6 uC/(4 pi eps d^2)
The problem with that is that those signs don't mean anything.

What you should have done:
Ea = 18 uC/(4 pi eps d^2) (towards c)
Ec = 6 uC/(4 pi eps d^2) (towards c)

Since they both point in the same direction, just add them up: Etotal = (18 + 6)uC/(4 pi eps d^2) (towards c)

Imagine how it would have been different if the charge at c were positive. In that case, Ec would have pointed away from c, the opposite to Ea. In that case you would have gotten Etotal = (18 - 6)uC/(4 pi eps d^2) (towards c)
 
  • #11
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Oh ok, I see. You're right, I'm sorry for getting so confused.
Oh well, even if I can't get points for it now, at least I'll know what to do when the test comes around. Thank you very much for being patient with me.
 
  • #12
what is epsilon here? what number would that be? what would i plug in for epsilon? and is d the distance? which would be r?
Ea = 18 uC/(4 pi eps d^2) (towards c)
Ec = 6 uC/(4 pi eps d^2) (towards c)
 
  • #13
Doc Al
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what is epsilon here? what number would that be? what would i plug in for epsilon?
ε0 is the permittivity of free space. See: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html" [Broken]
and is d the distance? which would be r?
Yes.
 
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