# Charges on a axis

1. Feb 3, 2012

### gotpink74

HELP I don't understand at all

1. The problem statement, all variables and given/known data
Two positive point charges on the y-axis, each of which has a charge of 5.0 10-9 C, are located at y = +0.60 m and y = -0.60 m. Find the magnitude and direction of the resultant electric force acting on a charge of 7.6 10-9 C located at x = 0.80 m on the x-axis.

2. Relevant equations

dont know

3. The attempt at a solution

I have tired to many ways with to many numbers.
My teacher hasn't taught these!

2. Feb 3, 2012

### SammyS

Staff Emeritus
Re: HELP I don't understand at all

Some teachers are like that, however, a few of my students said the same thing about me.

Find the force on the 7.6 nC charge due to each of the other charges, individually.

3. Feb 3, 2012

### gotpink74

Re: HELP I don't understand at all

so I do 7.6e-9*9e9/70^2

4. Feb 3, 2012

### gotpink74

1. The problem statement, all variables and given/known data
Two positive point charges on the y-axis, each of which has a charge of 5.0 10-9 C, are located at y = +0.60 m and y = -0.60 m. Find the magnitude and direction of the resultant electric force acting on a charge of 7.6 10-9 C located at x = 0.80 m on the x-axis.

2. Relevant equations
F=kq1q2/r^2

3. The attempt at a solution
I don't even know where to start my teacher hasn't taught us this. I have tried to do a vector diagram but it didn't work.

5. Feb 3, 2012

### Staff: Mentor

Can you show us your diagram? You will get two force vectors acting on the 3rd charge, and the net force is just the componen-wise sum of those forces...

6. Feb 3, 2012

### gotpink74

I attached my picture down below from their I do not know where to go

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7. Feb 3, 2012

### Staff: Mentor

Good start. Now draw the two force vectors -- one each from the charges on the y-axis. You draw the force vector arrows at the test charge, pointed along the line between the charges. So for the force on the test charge from the top y-axis charge, you draw the force vector at the test charge along the extended line between those two charges. Whether you draw the force vector arrow pointing back up at the top charge, or pointing away down and right, depends on whether the force is attractive or repulsive. Opposite charges attract, right?

Then once you have the two force vector arrows at the right test charge, divide them up into their x & y components, add components, and then combine those two final components to get the total force vector.

8. Feb 3, 2012

### gotpink74

what are X Y components?

9. Feb 3, 2012

### Staff: Mentor

10. Feb 3, 2012

### gotpink74

is the x y (40,70)

is this how you find the magnitude

11. Feb 3, 2012

### Staff: Mentor

Actually I now notice that your diagram does not match the problem definition. The problem definition mentions y= +/- 0.60m and x = 0.80m. Where did your 40 and 70 numbers come from in your diagram?

Once you get the dimensions right on the figure, draw your two force vector arrows, and use basic trig to get the x and y components of each vector. Then add the components to get the components of the total resultant force.

12. Feb 3, 2012

### gotpink74

is the hypotenuse 100

13. Feb 3, 2012

### SammyS

Staff Emeritus
Re: HELP I don't understand at all

Where does the 70 come from?

What is Coulomb's Law for electric force?

14. Feb 3, 2012

### gotpink74

Re: HELP I don't understand at all

Do I do √0.60^2+0.80^2

15. Feb 3, 2012

### gotpink74

does anyone know how to do this

16. Feb 3, 2012

### gotpink74

Re: HELP I don't understand at all

Im just trying to get the most help I can

17. Feb 4, 2012

### SammyS

Staff Emeritus
Re: HELP I don't understand at all

Certainly this quantity is involved in this problem.
$\displaystyle\sqrt{0.60^2+0.80^2}$ is the distance (in meters) from the 7.6 nC charge to either of the 5 nC charges.​
What is the force on the 7.6 nC charge due to the charge at y = +0.60 m ?