Calculating Force Vectors Between Charged Particles

[Mushroom79]

Homework Statement

http://imgur.com/opD71

Q1 = -2 C
Q2 = 3 C

Decide the forcevector F

Homework Equations

F = (k*Q1*Q2)/r^2

k = 8,99*10^9 Nm^2/C^2

The Attempt at a Solution

Q1, Q2 = squarerot(5+7) = 9,24 mm = 0,924 cm = 0,00924 m

F = (8,99*10^9*-2*3)/0,00924 = -5,838 * 10^12 N

Uniformity gives

F/Fx = 0,924/0,7

Fx = F/(0,924/0,7) = -4.423 N

F/Fy = 0,924/0,5

Fy = F/(0,924/0,5) = -3,159 N

F = (-4,423 N, -3,159 N) ?

 

[ap123]

A few questions for you :
Have you drawn a diagram for this with the direction of the forces on each charge?
What force are you trying to calculate?
What are the positions (or separation) of the 2 charges?

 

[gneill]

Homework Statement

http://imgur.com/opD71

For some reason your image is not showing up, so here's a copy:

Q1 = -2 C
Q2 = 3 C

Decide the forcevector F

Question: Which force vector do you want? The one acting on Q1 or the one acting on Q2? In other words, F₂₁ or F₁₂?

Homework Equations

F = (k*Q1*Q2)/r^2

k = 8,99*10^9 Nm^2/C^2

The Attempt at a Solution

Q1, Q2 = squarerot(5+7) = 9,24 mm = 0,924 cm = 0,00924

That distance doesn't look right. Better check it.

F = (8,99*10^9*-2*3)/0,00924 = -5,838 * 10^12 N

Uniformity gives

F/Fx = 0,924/0,7

Fx = F/(0,924/0,7) = -4.423 N

F/Fy = 0,924/0,5

Fy = F/(0,924/0,5) = -3,159 N

F = (-4,423 N, -3,159 N) ?

It looks like you're losing a good number of powers of ten from the force. The components should be around the same order of magnitude as the force they add up to; your F (currently) has an order of magnitude of 10¹².

 

[Mushroom79]

Updated picture with direction of the force vector I need (acting on Q2):

http://imgur.com/c7hu8

I'm trying to calculate the vectorforce F

Q1, Q2 = squarerot(5^2+7^2) = 8,602 mm would be the distance between the 2 charges

 

[ap123]

You have to decide which force to calculate, the force on Q1 or Q2?
Q2 is the easiest.
Start by working out the magnitude of the force first and then you can do the components.

 

[gneill]

Updated picture with direction of the force vector I need (acting on Q2):

http://imgur.com/c7hu8

I'm trying to calculate the vectorforce F

Vectors have a direction. While the forces acting on the two charges will have the same magnitude, their directions will be opposite -- equal and opposite forces thanks to Newton's third law. So you need to decide which force vector you want to calculate (they differ only in direction).

Here's a picture to make things clear:

Q1, Q2 = squarerot(5^2+7^2) = 8,602 mm would be the distance between the 2 charges

That looks better.

EDIT: Added diagram.

 

[Mushroom79]

Okay, let's say I want to calculate the force on Q2.
To get the magnitude (length) isn't it the same as F = squarerot(5^2+7^2) = 8,602 mm ?

After that I use the uniformity F/Fx = F/0,7

F/Fy = F/0,5

Not sure how I know what axis (x or y) should be negative or positive?

 

[ap123]

The magnitude of the force is given by Coulomb's law.
F = (k*Q1*Q2)/r^2

You can see the direction of the force vector on gneill's diagram.
If your x-axis is to the right, and your y-axis is straight up, then both components will be positive.
You can use your "uniformity" expressions will a small change :
F/Fx = 8.6/7
F/Fy = 8.6/5

( or you can use trigonometry - it'll come out to the same ).

 

[Mushroom79]

The magnitude of the force is given by Coulomb's law.
F = (k*Q1*Q2)/r^2

You can see the direction of the force vector on gneill's diagram.
If your x-axis is to the right, and your y-axis is straight up, then both components will be positive.
You can use your "uniformity" expressions will a small change :
F/Fx = 8.6/7
F/Fy = 8.6/5

( or you can use trigonometry - it'll come out to the same ).

I see, so:

F = -7,290*10^-14


F/Fx = 8,6/0,7

Fx = (-7,290*10^-14)/(8,6/0,7) = -5,934*10^-15 N

F/Fy = 8,6/0,5

Fy = (-7,290*10^-14)/(8,6/0,5) = -4,238*10^-15 N

F = (5,934*10^-15 N, 4,238*10^-15 N )

 

[ap123]

Almost there :)

F = 7.290*10^14 (not -14)

 

[Mushroom79]

Almost there :)

F = 7.290*10^14 (not -14)

Ah, I may be a little too fast sometimes.

Thank you both for great help