# Help with Coulombs Law

1. Jan 23, 2012

### gotpink74

1. The problem statement, all variables and given/known data
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

2. Relevant equations
F=kQ1Q2/r^2

3. The attempt at a solution

which way is counterclockwise

2. Jan 23, 2012

### Crwth

If it's counterclockwise from the negative x-axis, start from the left x-axis then go down.

3. Jan 23, 2012

### SammyS

Staff Emeritus
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.

4. Jan 23, 2012

### gotpink74

What is the net electric field at the position of the charge at the origin? How do you find this

5. Jan 23, 2012

### Crwth

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

6. Jan 23, 2012

### gotpink74

does the nC stand for 10^-9

7. Jan 23, 2012

### Crwth

Yes, the n stands for nano, $10^{-9}$

8. Jan 23, 2012

### gotpink74

how do you find part a.

9. Jan 23, 2012

### Crwth

Have you attempted to solve it?

10. Jan 23, 2012

### gotpink74

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

11. Jan 23, 2012

### Crwth

12. Jan 23, 2012

### gotpink74

(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

13. Jan 23, 2012

### gotpink74

14. Jan 23, 2012

### Crwth

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.

15. Jan 23, 2012

### gotpink74

do i need to then find the hypotunse

16. Jan 23, 2012

### gotpink74

i dont understand why I need a counterclockwise one arent they the same

17. Jan 23, 2012

### Crwth

You want to find the magnitude, and the direction (angle)

18. Jan 23, 2012

### gotpink74

Nothing has helped

19. Jan 23, 2012

### gotpink74

how do i find the direction angle I was given no number for an angle

20. Jan 23, 2012

### Crwth

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.