# Vector addition in electric field

1. Jan 12, 2008

### hteezy

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

Two positive charges = = 2.0 are located at = 0, = 0.30 and = 0, = -0.30 , respectively. Third point charge = 4.0 is located at = 0.40 , = 0.What is the net force ((a)magnitude and (b)direction) on charge exerted by the other two charges?

2. Relevant equations

Columbs law (force between two point charges)
F = 1/4 piEo times abs value of q1*q2 divided by r^2

3. The attempt at a solution

first i have to find the force that Q exerts on q1 (F1 on q1)
i use coulumbs law
F = k * [q1q2]/[r^2]
k = 8.988 e 9

so i plug in q1 and Q and the distance between them and i get....

F1 on q1 = .29
i break that down into components x and y

F1x= .29cos(.40/.50) = .287
F1y = .29sin(.30/.50) = .00301

so now i find the force that q2 exerted on q1 (F2 on q1)

q2 and q1 are both on the vertical axis so there will only be a y component

so i use cuolumbs law again and now i plug in q1 and q2 and r is the distance between them so..

F2= k [2.0 e -6][2.0 e -6] / (.60 ^2) = .1

so now i add the x components and y components up

there is only one x component and 2 y components sooo

F1x = .287
F1y + F2y = .10301

F = square root of ( .287^2 + .10301^2)

which is equal to .30 (2 sig figs)

then to find the angle i would use tan(theta) = Fy/Fx
which gives me an angle of 19.7 degrees
It is supposed to be in the +x direction so i think i am supposed to subtract it from 180 and that should give me my angle

im afraid to type the answer in into masterinphysics because i have only a few attempts left...so i want to make sure im right before i do anything!

2. Jan 13, 2008

### Staff: Mentor

It's hard to understand what you've written (when I compare it to what follows). Please fill in the gaps. What are the charges? Where are they? (I assume those are x and y coordinates?) You need the force on which charge?

OK.

What are the charges?
Redo these calculations. Also: Direction (and thus signs) counts.

3. Jan 13, 2008

### hteezy

redo

opps sorry the question didnt copy right. my fault

Two positive charges q1 = q2 = 2.0 $$\mu$$C are located at coordinate (0,.30) meters and (0,-.30) meters, respectively. The third point charge Q = 4.0 $$\mu$$C and is located at (.40,0) meters. What is the net force magnitude and direction on chare q1 exerted by the other two charges.

i made a lil drawing so that it could help give a visual...its attached

So first i found the force exerted by Q on q1 which i will call F1
the diff between q1 and Q is .50 meters

F1 on q1 = k (2.0 e -6)(4.0 e -6) / (.50^2) = .29 N

i then had to break that down into x and y components

i need to use cosine for the x component so... = F1 * cos (.40/.50)
and then the y component would be = F1 * sin (.30/.50)

i hope that makes sense now

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4. Jan 14, 2008

### Staff: Mentor

Much better.
Good. That's the magnitude of the force.

F1*cos(theta) = F1*(.40/.50)
F1*sin(theta) = F1*(.30/.50)

Those are the magnitudes of the components. What are their signs?

5. Jan 14, 2008

### hteezy

signs?

I guess they are positive?

6. Jan 14, 2008

### Staff: Mentor

Draw yourself a picture of the force that Q exerts on q1. Which way does the vector point?

7. Jan 14, 2008

### hteezy

ok so the force that Q exerts on q1 is negative
and the force that q2 exerts on q1 is positive

8. Jan 14, 2008

### hteezy

so F1 and its components are negative?

9. Jan 14, 2008

### hteezy

actually no...only the x component is in the negative direction for F1

the y component of F1 is still positive

10. Jan 14, 2008

### Staff: Mentor

I don't understand what you mean by saying that the force is negative or positive.

F1 is shown on your diagram in post #3. Which way does its x-component point? Its y-component?

11. Jan 14, 2008

### hteezy

its x component points in the - x direction
the y comp. points in the + y direction

12. Jan 15, 2008

### Staff: Mentor

Right!

13. Jan 15, 2008

### hteezy

ok and the force exerted by q2 on q1 (F2) is also in the positve y direction
and there is no x component for that force

so now that i know the components for all the forces do i just add them all up...add the y component of F1 and F2 together?

then do i just add the final x and y components together?

and that would be the magnitude of the force exerted on q1 by the other two charges right?

14. Jan 15, 2008

### Staff: Mentor

Good.
Yes.
No. The x and y components are perpendicular to each other--so you can't just add them like numbers. How do you find the magnitude (and direction) of a vector, given its components?

15. Jan 15, 2008

### hteezy

F = $$\sqrt{}(Fx^2 + Fy^2)$$

ok i think i got it...Thank you very much for your help!!!!!