# Homework Help: Electrostatic Force question

1. Sep 2, 2007

### itsagulati

1. The problem statement, all variables and given/known data
Three charges are fixed to an x, y coordinate system. A charge of +18 µC is on the y axis at y = +3.0 m. A charge of -16 µC is at the origin. Lastly, a charge of +45 µC is on the x axis at x = +3.0 m. Determine the magnitude and direction of the net electrostatic force on the charge at x = +3.0 m. Specify the direction relative to the -x axis.

2. Relevant equations

F = k*q1*q2 / r^2

3. The attempt at a solution

9e9 * 18e-6 * 16e-6 / 3^2 = .28768 N

and

9e9 * 45e-6 * 16e-6 / 3^2 = .7192N

(.29^2 + .72^2)^.5 = .7746N

tan^-1 ( .29/.72) = 21.8 degrees...90 - 22 = 68 degrees below the x-axis

Can anyone figure out where I am going wrong? I don't get what i am not seeing.

Thank you!

2. Sep 3, 2007

### learningphysics

This is wrong for calculating the net force... first find the net x-component of the force acting on the charge (so add the two components due to each charge)... then find the net y-component (only one charge needs to be considered here since the charge on the x-axis only contributes an x-component)...

Then use the pythogrean theorem...

3. Sep 3, 2007

### itsagulati

If they are straight on the x-axis just out 3.0m and the other one straight up on the y-axis just up 3.0 m...then doesn't that make them 90 degree angles? meaning no x component on the charge on the y axis and then for the charge on the x-axis there is no y component?

So, just a charge 3.0 m out at 0 degrees and one 3.0m up at 90 degrees?
Thus only the .29 and .72 (clearly it is wrong, but i am not getting how to calculate otherwise).

Thanks!

4. Sep 3, 2007

### learningphysics

The charge at (0,3) exerts both an x and y component on the charge at (3,0)...

The question is asking for the force on the charge at (3,0)... not the force on the charge at (0,0)

5. Sep 3, 2007

### itsagulati

Ok so i was just not htinking in vectors...but here is what i have (i only have one more chance at this before it is due so i am trying to be cautious).

K * 18e-6 * 16e-6 / 9 = .288 = y comp1
K* 18e-6*45e-6/9 = .81 = x comp1

K*45e-6*18e-6/9 = .81 = y comp 2
K*45e-6*16e-6/9 = .72 = x comp2

x1 + x2 = 1.53N
y1+y2 = 1.098 N

(1.098^2 + 1.53^2)^.5 = 1.88N

THEN for the angle i am doing tan^-1 (1.098/1.53) = 35.66 degrees and relative to the -x axis that is -35.66 degrees if you draw the line to extend back through the origin and into quadrant 3?

And thanks for all of your help so far :)
It is definitely helping me past the road blocks!

6. Sep 3, 2007

### learningphysics

You're making a common mistake calculating the components... Please look at post #22 in this thread:

Also, I'm confused as to why you have a calculation for a force between the 18muC and -16muC charges... you only need the force on the 45muC charge...

The force that the -16muC charge exerts on the 45muC charge is 0.72N in the -x direction... or -0.72 in the x direction... that part is right.

Now you need to calculate the x and y components of the charge by the 18muC on the 45muC after looking at the post in the thread I linked...

First calculate the net force of the 18muC on the 45muC. Then calculate the x component and y component.

7. Sep 3, 2007

### itsagulati

ok...it asks for: kq / x^2+y^2 * cos @ (where @ = theta).

only thing i can think of is 45 degrees...so:

kq / x^2 + y^2 * cos (45) = ((9e9 * 16e-6) / (3^2))* cos (45) + ((9e9 * 18e-6) / (3^2))* cos (45)?

(currently working on this...i realize i forgot the x and y component part of the problem)

Last edited: Sep 3, 2007
8. Sep 3, 2007

### learningphysics

No, the x^2 + y^2 is in the denominator...

What is the net force of the 18muC on the 45muC... what is r? what is r^2?

9. Sep 3, 2007

### itsagulati

bleh. sorry the problems are getting jumbled in my head...

F = 9e9 * 45e-6 * 18e-6 / 9 = .81 N
and we already determined .72 N

So, now i calculate the x and y component?

i get: x1 = 0, y1 = .81
x2 = .72, y2 = 0

pyth. theo. => (.81^2+.72^2)^.5 = 1.08N?

10. Sep 3, 2007

### learningphysics

Why do you have 9 in the denominator?

11. Sep 3, 2007

### itsagulati

God i don't get this problem...

i got all the starred, double starred, etc. problems...and this no star basic arse problem i just am not getting. Why on earth is this one so hard for me LOL

F***

Last edited: Sep 3, 2007
12. Sep 3, 2007

### itsagulati

9 in the denominator is from 3^2 as r^2

the only other thing i have come up with is since it is the repulsive force...then 3 sqrt(2) so 18 in the denominator giving me .405

.405 and .72 are my two sides then?

13. Sep 3, 2007

### learningphysics

Yes, it should be 18 in the denominator... I'm not understanding why you say "since it is the repulsive force"... whether it was attractive or repulsive... the distance is 3sqrt(2) hence r^2 = 18... Hope I don't seem like a jerk... just want to make sure you understand this part.

No wait... 0.405 N is the net force exerted by the charge at (0,3) on the charge at (3,0)... that's right... now what is the x-component of this force... and what's the y-component of this force...