Magnitude of electric field and 3 points

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Homework Statement



Points A, B, and C are at the vertices of an equilateral triangle. A certain positive charge q placed at A produces an electric field of magnitude 145 N/C at C. Suppose a second, identical charge is placed at B. What is the magnitude of the new electric field at C?


Homework Equations



Not sure what to use... ?

E=F/q'
or
F=q'E



The Attempt at a Solution



I did not know where to start..

Thank you.

Thanks again to anyone who replies. It is late here so I am signing off.. I will reply tomorrow if anyone has responded.
 
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Answers and Replies

  • #2
gneill
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You have the magnitude of the field produced by the charge placed at A, and the geometry of the layout gives you its direction. So you can deduce a vector to represent the field at C produced by the charge at A.

Now you have an identical charge placed at B. Because of the symmetry of the layout, what can you say about the magnitude and direction of the field that this new charge will produce at C?
 
  • #3
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You have the magnitude of the field produced by the charge placed at A, and the geometry of the layout gives you its direction. So you can deduce a vector to represent the field at C produced by the charge at A.

Now you have an identical charge placed at B. Because of the symmetry of the layout, what can you say about the magnitude and direction of the field that this new charge will produce at C?

I thought maybe that the direction from B would be towards C just like the direction from A is to C. I am not sure if I am looking at that right though. Wouldn't the magnitude be the same??
 
  • #4
gneill
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I thought maybe that the direction from B would be towards C just like the direction from A is to C. I am not sure if I am looking at that right though. Wouldn't the magnitude be the same??

It's an equilateral triangle. The side lengths are equal. The interior angles are all equal (what are they?) So this distance AC is equal to the distance BC. The only difference is the direction. What law applies to determine the magnitude of the forces? What can you say about the directions of the forces?
 
  • #5
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It's an equilateral triangle. The side lengths are equal. The interior angles are all equal (what are they?) So this distance AC is equal to the distance BC. The only difference is the direction. What law applies to determine the magnitude of the forces? What can you say about the directions of the forces?

They are 60 degrees. Coulomb's Law, the forces will be opposite?
 
  • #6
gneill
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Why opposite directions? State your reasoning.

Draw the diagram. Electric field directions are always radial from point charges; they lie along the line joining the "test" location to the charge. The direction of the field along the line depends upon the sign of the charge.
 
  • #7
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Why opposite directions? State your reasoning.

Draw the diagram. Electric field directions are always radial from point charges; they lie along the line joining the "test" location to the charge. The direction of the field along the line depends upon the sign of the charge.

I was thinking positive repels positive and maybe it would push the field away. Now I am thinking they will both go towards point c so I have to figure both going in the direction towards point C.
 
  • #8
gneill
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Have a look http://www.physicsclassroom.com/class/estatics/u8l4c.cfm" [Broken].
 
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  • #9
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Have a look http://www.physicsclassroom.com/class/estatics/u8l4c.cfm" [Broken].

Thanks! I will review that material and if I have more trouble, I will let you know. Thanks so much for your time!
 
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