1. Sep 12, 2007

### WardenOfTheMint

1. The problem statement, all variables and given/known data
Determine the electric field strength at a point 0.01m to the left of the middle charge.

2. Relevant equations

Electric Field
E = Ke (q / (r^2) )

Electric Force
F = Ke (q1q2 / r^2)

Constant
Ke = 9.00e9 Nm^2/C^2

3. The attempt at a solution

http://img504.imageshack.us/img504/5137/myillustrationql5.jpg [Broken]

Let 7e-6C be q1, 1.0e-6 q2, and -3e-6C q3.
So if you start at the middle charge and go 0.01m to the left, I imagine a point P, and the distance between the middle charge and point P should be 0.01m I think...

So from there...

E q1 on P = 9e9 Nm^2/C^2 ( 7e-6C / 0.03m^2 ) = 7e7 N / C.

E q2 on P = 9e9 Nm^2/C^2 ( 1e-6C / 0.01m^2 ) = 9e7 N / C.

E q3 on P = 9e9 Nm^2/C^2 ( -3e-6C / 0.03m^2 ) = -3e7 N / C. 0.03m because 0.01m+0.02m I think.

Enet = 7e7 + 9e7 - 3e7 = 1.3e8 N/C.

But that's wrong!

Then you have to find the magnitude (in Newtons) if a charge of -3e-6C is placed at this imaginary point.

Last edited by a moderator: May 3, 2017
2. Sep 12, 2007

### learningphysics

Your signs are wrong... take right as positive, left as negative... what is the direction of the field due to the first charge... what is the direction of the field due to the second charge... and the third...

3. Sep 12, 2007

### WardenOfTheMint

first thank learnphysc, you are always around!!!!

i'm not getting it. E q1 on p should be positive shouldn't it? q1 is acting on point P forcing it to move right?

4. Sep 12, 2007

### learningphysics

no prob. Yes, but what about q2 and q3?

5. Sep 12, 2007

### WardenOfTheMint

oh yea. so E q2 on P should be negative because it's being to the left. ok i got it. does the negative charge push it also push it to the left?

6. Sep 12, 2007

### WardenOfTheMint

so maybe the answer should be 7e7 N / C + ( -9e7 N / C) + ( -3e7 N / C) = -5e7N/C
i don't know the answer so i can only hope it's right

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

### learningphysics

no the negative charge creates a field to the right... so it's positive...

If you're not sure of the direction of the field use a positive test charge in the place where you're trying to get the field... so the direction of the force acting on a positive test charge, is the direction of the field...

So the negative charge would attract a positive test charge towards the right... so it creates a positive field at that point...

8. Sep 12, 2007

### learningphysics

should be +3e7 N/C for the negative charge.

9. Sep 12, 2007

### WardenOfTheMint

so in electric field questions a positive charge will always "push" and the negative will do the "pulling'?

10. Sep 12, 2007

### WardenOfTheMint

thanks! i got a confirmation it's right 7e7 N / C + ( -9e7 N / C) + ( 3e7 N / C) IS RIGHT!!! woohoo

11. Sep 12, 2007

### learningphysics

Yes, in electric field questions it is like that since the test charge is positive. but not in electric force though.

12. Sep 12, 2007

### learningphysics

Cool!

13. Sep 12, 2007

### BlackWyvern

E = kq/d^2

Just add up all of the three vectors you get.