Determining Electric Field at Origin Due to Charges A and B

In summary, the problem asks to determine the electric field at the origin due to two charges at A and B, with charge B being reversed in sign. The charges form an equilateral triangle with the origin. By using the equation E=k*Q/r^2 and the fact that the angles of an equilateral triangle are equal, the components can be solved in terms of q and L. After some algebra and vector addition, the final answer is the square root of 3 times k*q/L^2.
  • #1
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Homework Statement


Determine the electric field E at the origin 0 in fig 16-56 due to the two charges at A and B. b.) Repeat, but let the charge at B be reversed in sign. I can;t find the figure online, so I'll have to describe it. Charge A is on the vertical axis distance L from the origin with charge B to the lower right at a distance of L which is also a distance of L from the origin. The origin, Charge A, and Charge B form an equilateral triangle. Charge A and B for the first part are both +Q.

Homework Equations


E=k*Q/r^2
tan^-1(y/x) = theta

The Attempt at a Solution


I understand that the electric field has to be calculated then the components added as vectors, and I understand how the answer in the back of the book was set up, except that the answer is the square root of 3 times k*q/L^2. I don't see how they could have gotten the square root of three with so little detail in the problem, but I'm assuming it has something to do with the angle between them? I've finished all my other more advanced problems but I'm still stumped by this one. Once I get the components I'll be able to calculate the angle, but I'm wondering how they got what they did. Thanks, all help is appreciated! Again, sorry for lack of diagram.
 
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  • #2
Notice that they tell you that the charges form an equilateral triangle with the origin. What are the angles of an equilateral triangle equal to?

Assuming I am picturing this correctly, you should be able to use this information to get the components in terms of just q and L.
 
  • #3
Thanks, but I got it in school today. Wasn't very hard, but took a bit of algebra and vector addition. Thanks anyway!
 

FAQ: Determining Electric Field at Origin Due to Charges A and B

1. What is an electric field?

An electric field is a physical field that surrounds a charged particle and exerts a force on other charged particles within its vicinity. It is a vector quantity, meaning it has both magnitude and direction.

2. How is the electric field determined?

The electric field is determined by measuring the force exerted on a test charge placed in the field. The magnitude of the electric field is equal to the force divided by the charge of the test particle.

3. What factors affect the strength of an electric field?

The strength of an electric field is affected by the magnitude of the charge creating the field, as well as the distance between the charged particle and the point at which the field is being measured. The direction of the field is also affected by the sign of the charge.

4. Can the electric field be negative?

Yes, the electric field can have a negative value. This occurs when the source charge and the test charge have opposite signs, resulting in a force directed in the opposite direction to the field.

5. What is the SI unit of electric field?

The SI unit of electric field is newtons per coulomb (N/C). This represents the force exerted on one coulomb of charge in the field.

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