# Homework Help: Force on a point from electric charges

1. Apr 3, 2013

### rocapp

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

What is the net force on the bottom charge?
Please give the force in Fx and Fy
2. Relevant equations

F=qE
E=kq/r^2
3. The attempt at a solution
Fx=2.88x10^-9N
Fy=-1.01x10^-4N

These aren't correct. I used the equation above and used

r=0.035m for both the side charges.

For the top charge,

r=0.05m

Why is the formula not working?

Thanks!

2. Apr 3, 2013

### ehild

Why do you use r=3.5 cm for the side charges when r=5 cm is indicated?

ehild

3. Apr 4, 2013

### rocapp

I have to separate the forces into their x and y components, so the distance on the x axis should be 3.5 cm. is this not the right approach?

4. Apr 4, 2013

### ehild

How do you use the distance on the x axis at all? Show your working in detail.

ehild

5. Apr 4, 2013

### tiny-tim

hi rocapp!
nooo

you want Q/r2 times cosθ,

but you're using Q/(rcosθ)2

6. Apr 4, 2013

### rocapp

Ohh! Thanks so much!

7. Apr 4, 2013

### rocapp

Does the 1nC charge affect itself in terms of force?

I keep getting wrong answers still.

-2.124×10^-23 N is the y-component I get from the -6nC charge.
1.00x10^-23 N is the y-component I get from both 2 nC charges on the 1 nC charge.
Fy = -1.124x10^-24 N

But that's not correct.

Fx=1.00x10^-23 N

Not sure if either of these are correct.

Last edited: Apr 4, 2013
8. Apr 4, 2013

### ehild

How did you get these data? Show your work in detail, please.

ehild

9. Apr 4, 2013

### rocapp

In the images attached are the exact equations I used for Fx and Fy.

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• ###### MSP50951a9a2ef6392c2hda00002fa3dge1i579dcc5.gif
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10. Apr 4, 2013

### ehild

You use wrong value for k in Coulomb's Law.

You need the resultant force on the bottom charge. Fx should be the horizontal component of the resultant force, Fy has to be the vertical component. What are the directions of the force components exerted bas the charges?

ehild

Last edited: Apr 5, 2013
11. Apr 5, 2013

### rocapp

Ah. Thanks for that!

The horizontal x component of the -6 nC charge is zero because it is directly above the 1 nC charge.

The horizontal x component of one of the 2 nC charges is:

Fx = (k*q*q/r^2)*cos(theta)
Fx = ((8.99x10^9)*(2x10^-9)*(1.0x10^-9)/(.05^2))*cos(45)
Fx = 5.09x10^-6 N

Since there are two, the total Fx = 2*5.09x10^-6 N = 1.02x10^-5 N

The y-components of the two 2 nC charges are the same as the x components because sin(45)=cos(45).

The y-component of the -6 nC charge is

Fy=((8.99x10^9)*(1.0x10^-9)*(-6x10^-9)/(0.05^2))
Fy= -2.16x10^-5 N

So the total Fy= Fy + Fy = (-2.16x10^-9) + (1.02x10^-5) = 1.02x10^-5 N

This is still not correct, though.

12. Apr 5, 2013

### ehild

Check the direction of these forces. Are not the x components opposite to each other?

Show what you mean on positive y direction, up of down?

Do not note the different force components with the same Fy. And you made a mistake, it is not 2.16x10^-9 but 2.16x10^-5.

ehild

13. Apr 5, 2013

Thanks!