Magnitude of electric field and 3 points

In summary, two identical charges placed at the vertices A and B of an equilateral triangle will produce opposite electric fields at the third vertex C, with the same magnitude of 145 N/C. This is due to Coulomb's Law and the radial nature of electric fields from point charges.
  • #1
ENCgirl
12
0

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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  • #2
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
gneill said:
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
ENCgirl said:
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
gneill said:
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
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
gneill said:
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
Have a look http://www.physicsclassroom.com/class/estatics/u8l4c.cfm" .
 
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  • #9
gneill said:
Have a look http://www.physicsclassroom.com/class/estatics/u8l4c.cfm" .

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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FAQ: Magnitude of electric field and 3 points

What is the magnitude of electric field?

The magnitude of electric field is a measure of the strength of an electric field at a specific point. It is represented by the symbol E and is measured in units of volts per meter (V/m).

How is the magnitude of electric field calculated?

The magnitude of electric field is calculated by dividing the electric force acting on a test charge by the magnitude of the test charge itself. This can be represented by the equation E = F/q, where E is the electric field, F is the electric force, and q is the test charge.

What factors affect the magnitude of electric field?

The magnitude of electric field is affected by the distance from the source of the electric field, the amount of charge on the source, and the medium through which the electric field is passing. Additionally, the presence of other charges in the vicinity can also affect the magnitude of electric field.

What is the significance of the direction of electric field?

The direction of electric field is important as it indicates the direction in which a positive test charge would move if placed in that field. The direction of electric field is always directed away from positive charges and towards negative charges.

What are some applications of understanding the magnitude of electric field?

Understanding the magnitude of electric field is important in various fields such as electrical engineering, physics, and telecommunications. It is used to design and analyze electrical circuits, calculate the forces acting on charged particles, and to study the behavior of electromagnetic waves, among other applications.

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