# Homework Help: Electric field on point charge

1. Sep 26, 2008

### gnarkil

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

what is the electric field at point A (0,12) if there is charge c of 8*10^-9 coulombs at (16,0) and a charge d of 6*10^-9 coulombs at (-9,0)
determine x and y components of electric field (see diagram)

2. Relevant equations

electric field e = kq/r^2 where k is 9*10^9, q is charge, r is distance from point

3. The attempt at a solution

x components:
e_c = [(9*10^9)(8*10^-9)/(sqrt(16^2 + 12^2)^2 ]sin(phi)
= [(9*10^9)(8*10^-9)/(sqrt(16^2 + 12^2)^2 ] * [ 16 /(sqrt(16^2 + 12^2))]
= [(9*10^9)(8*10^-9)(16)] / [(16^2 + 12^2)^(3/2)]
= 0.144 newtons/coulomb

e_d = [(9*10^9)(6*10^-9)/(sqrt(9^2 + 12^2)^2 ]sin(theta)
= [(9*10^9)(6*10^-9)/(sqrt(9^2 + 12^2)^2 ] * [ 9 /(sqrt(9^2 + 12^2))]
= [(9*10^9)(6*10^-9)(9)] / [(9^2 + 12^2)^(3/2)]
= 0.144 newtons/coulomb

y components:
e_c = [(9*10^9)(8*10^-9)/(sqrt(16^2 + 12^2)^2 ]cos(phi)
= [(9*10^9)(8*10^-9)/(sqrt(16^2 + 12^2)^2 ] * [ 12 /(sqrt(16^2 + 12^2))]
= [(9*10^9)(8*10^-9)(12)] / [(16^2 + 12^2)^(3/2)]
= 0.108 newtons/coulomb

e_d = [(9*10^9)(6*10^-9)/(sqrt(9^2 + 12^2)^2 ]cos(theta)
= [(9*10^9)(6*10^-9)/(sqrt(9^2 + 12^2)^2 ] * [ 12 /(sqrt(9^2 + 12^2))]
= [(9*10^9)(6*10^-9)(12)] / [(9^2 + 12^2)^(3/2)]
= 0.192 newtons/coulomb

net electric force components:
x = 0.144+0.144 = 0.288 newtons/coulomb
y = 0.108+0.192 = 0.300 newtons/coulomb

my net components are wrong. why?

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2. Sep 26, 2008

### Nabeshin

Your work is a little strange here in that you first find the vectors, and then component-ize them. I can't see your picture so I'm thinking you might have gotten an angle wrong.

You could try simply solving for the compoents without the general vector by saying in the y component both charges are 12 units away and in the x they are 16 and 9 respectively. This should help to minimize mistakes, just make sure to keep track of signs.

3. Sep 27, 2008

### gnarkil

i don't think i got the angles wrong, if you look at the diagram, is the way i calculated them the wrong way? will it affect the final answer?

4. Sep 27, 2008

### hage567

One thing I notice is that you added your x components. I think the x component due to charge C and the x component due to charge D are NOT pointing in the same direction. I don't know if that makes it right but I just thought I'd mention it so you could check it out.

5. Sep 27, 2008

### gnarkil

it seems both my net components are wrong. isn't the problem asking for the net components, therefore wouldn't i just add them regardless of direction?

6. Sep 27, 2008

### hage567

They are vectors, though. You are trying to find the net electric field at that point due to both charges. If the x component vectors act in opposite directions, you subtract them, they don't add up.

7. Sep 28, 2008

### gnarkil

so i should be subtracting vectors then? how do i know which vector component is the positive one, is it based on magnitude?

8. Sep 28, 2008

### scrplyr

using the vectors at point A -going upward (use A as the origin), the vector e_c is in the negative x-direction and positive y-direction. e_d is in the positive x, positive y. change your signs according to the directions, and that should change your answer for e_net(x). your net electric field in the y-direction is right.

9. Sep 28, 2008

### gnarkil

thanks for clarifying how to get the directions, i didn't even think to look at it that way, really cleared up things for me

so if i change the direction (signs) of my vectors and re-sum them i should get the answer i am looking for? are my original calculations, ignoring direction, correct?

so:

x components:
e_c = -0.144 newtons/coulomb
e_d = + 0.144 newtons/coulomb

y components:
e_c = + 0.108 newtons/coulomb
e_d = + 0.192 newtons/coulomb

is that correct now?

so net components
x = -0.144 + 0.144 = 0 newtons/coulomb
y = +0.108 + 0.192 = 0.300 newtons/coulomb

10. Sep 28, 2008

### scrplyr

yup, thats the right answer =)

11. Sep 29, 2008

### gnarkil

thanks it was right