Force on a point from electric charges

[rocapp]

Homework Statement

What is the net force on the bottom charge?
Please give the force in Fx and Fy

Homework Equations

F=qE
E=kq/r^2

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!

 

[ehild]

Homework Statement

What is the net force on the bottom charge?
Please give the force in Fx and Fy

Homework Equations

F=qE
E=kq/r^2

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!

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

ehild

 

[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?

 

[ehild]

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

ehild

 

[tiny-tim]

hi 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?

nooo …

you want Q/r² times cosθ,

but you're using Q/(rcosθ)²

 

[rocapp]

Ohh! Thanks so much!

 

[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.

 

[ehild]

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

ehild

 

[rocapp]

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

 

[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

 

[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.

 

[ehild]

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

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

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

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

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

This is still not correct, though.

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

 

[rocapp]

Thanks!