Three Point Charges, Find x-component

• rlc
In summary, the conversation discusses the arrangement of three point charges in an equilateral triangle and the determination of the direction and components of the force acting on one of the charges. The solution involves using Coulomb's law to find the magnitude of the force and then using the signs to determine the direction. It is important to properly square the distance in Coulomb's law to get the correct numerical values.
rlc

Homework Statement

In the figure, three point charges are arranged in a equilateral triangle of side a= 1.40×10-2 m. Sketch the field lines due to + Q and -Q, where Q = 2.50×10-5 C , and from them determine the direction of the force that acts on +q = 5.00×10-8 C because of the presence of the other two charges. What is the x-component of the force? What is the y-component of the force?

Homework Equations

F1=K*((Q*q)/a^2)
F(x)=2|F1|*Cos(60)

The Attempt at a Solution

I know that the y-component is 0 N.
Someone posted this problem in yahoo answers, and I tried to follow it. Something went wrong along the way, but here is the work from that solution:

*+Q=1, -Q=2, +q=3*

F1,3=(8.99E9)(5.00E-8)(2.50E-5)/(1.40E-2)=0.803 N
F2,3=(8.99E9)(5.00E-8)(-2.50E-5)/(1.40E-2)=-0.803 N

F1,3x=0.803(cos(30 degrees))=0.6951 (right)
F1,3y=0.803(cos(60))=0.4013 (down)

F2,3x=-0.803(cos(30))=-0.6951 (left)
F2,3Y=-0.803(cos(60))=-0.4013 (down

Here, I have the problem where if you combine x and y values, the answer is 0 for both of them.
Where did I go wrong, or is all of this wrong?

rlc said:
F2,3=(8.99E9)(5.00E-8)(-2.50E-5)/(1.40E-2)=-0.803 N
Get rid of that negative sign. Use Coulomb's law to find the magnitude of the force, then use the signs to find the direction. Draw yourself a diagram of the forces on q.

Sorry for not replying in a while!
In the work I did, I completely forgot to square a, making all my numbers wrong.

((9E9)(q)(+Q))/(a^2)
((9E9)(5.00E-8)(2.50E-5))/(1.40E-2)^2=57.33 N (this is the x-component)

Thank you!

1. What is the formula for finding the x-component of three point charges?

The formula for finding the x-component of three point charges is:

Fx = F1x + F2x + F3x, where Fx is the total x-component, F1x, F2x, and F3x are the x-components of each individual charge.

2. How do I determine the direction of the x-component?

The direction of the x-component can be determined by considering the location of the three point charges. If the charges are all on the same side of the x-axis, the x-component will be positive. If the charges are on opposite sides of the x-axis, the x-component will be negative.

3. Can the x-component of a charge be larger than the total force?

Yes, the x-component of a charge can be larger than the total force. This can occur when the x-components of the individual charges are in the same direction and add up to a larger value than the total force.

4. What happens to the x-component if one of the charges is negative?

If one of the charges is negative, the x-component will also be negative. This is because the negative charge will exert a force in the opposite direction of the positive charges, resulting in a negative x-component.

5. Can the x-component of a charge ever be zero?

Yes, the x-component of a charge can be zero. This can occur when the x-components of the individual charges cancel each other out, resulting in a total x-component of zero.

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