Coulomb's Law and Electric Fields for a Triangle of Charges

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gotpink74
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Homework Statement


Consider three charges q1 = 4.3 nC, q2 = 6.6 nC, and q3 = -2.3 nC, arranged in a triangle as shown below.

(a) What is the electric force acting on the charge at the origin?
N, ° counterclockwise from the negative x-axis

(b) What is the net electric field at the position of the charge at the origin?
N/C, ° counterclockwise from the negative x-axi
picture of problem
http://www.webassign.net/holtphys/p16-38alt.gif

Homework Equations


F=kQ1Q2/r^2


The Attempt at a Solution




which way is counterclockwise
 
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If it's counterclockwise from the negative x-axis, start from the left x-axis then go down.
 
gotpink74 said:

Homework Statement


Consider three charges q1 = 4.3 nC, q2 = 6.6 nC, and q3 = -2.3 nC, arranged in a triangle as shown below.

(a) What is the electric force acting on the charge at the origin?
N, ° counterclockwise from the negative x-axis

(b) What is the net electric field at the position of the charge at the origin?
N/C, ° counterclockwise from the negative x-axi
picture of problem
http://www.webassign.net/holtphys/p16-38alt.gif

Homework Equations


F=kQ1Q2/r^2

The Attempt at a Solution



which way is counterclockwise
Find a watch with a second hand. Set it at the origin, face up. The second hand has a clockwise rotation. the opposite rotation is counter-clockwise.
 
What is the net electric field at the position of the charge at the origin? How do you find this
 
You're essentially finding the force exerted by a charge on an infinitesimally small positive charge, which happens to be placed at the origin.

electric field = kQ/r^2
 
I have my last answers were
(a) What is the electric force acting on the charge at the origin?
2.838e-6 N, -8.901e-6 ° counterclockwise from the negative x-axis

(b) What is the net electric field at the position of the charge at the origin?
4.3e11 N/C, 3.87e12 ° counterclockwise from the negative x-axis BUT THEY WERE WRONG
 
(9*10^9)*(-2.3*10^(-9))*(4.3*10^(-9))/0.10^2

(9*10^9)*(4.3*10^(-9))*(6.6*10^(-9))/0.30^2 counterclockwise

net electric field
(9*10^9)*(4.3*10^(-9))/0.30^2
(9*10^9)*(4.3*10^(-9))/0.10^2
 
someone please help this is due tomorrow and I cannot figure it out!
 
For part a, you used Coulomb's law to find the force exerted by each charge on the charge at the origin. Have you paid attention to their directions? If you draw them as vectors, how would the resultant vector look?

And for electric field, you want to use the other charge as the Q you're using.
 
i don't understand why I need a counterclockwise one arent they the same
 
gotpink74 said:
i don't understand why I need a counterclockwise one arent they the same

You want to find the magnitude, and the direction (angle)
 
how do i find the direction angle I was given no number for an angle
 
You should have the two perpendicular force vectors starting at the origin; they form a right triangle... If you have the two legs of a right triangle, you should be fine figuring out the rest of the parts of the triangle.
 
i found the angles now i can't find the force and electric field on the orgin
 
You posted Coulomb's law for the electric force. Draw a diagram and use vectors to represent the each force on the charge at the origin, then you should be able to get the magnitude of the resultant vector.
 
The magnitude of the vector is always positive; you have the angle to account for the direction.

However, the actual vector can be going in the negative direction.