# Where is the Electric Field Equal to Zero?

1. Aug 24, 2007

### mrlucky0

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

Two tiny objects with equal charge of 85.0 µC are placed at two corners of a square with sides of 0.295 m, as shown. How far above and to the left of the corner of the square labeled A would you place a third small object with the same charge so that the electric field is zero at A?

Here is a drawing:

http://img177.imageshack.us/my.php?image=newbitmapimageqn1.png

2. Relevant equations

Coulomb's Law: F = (k * q1 * q2 ) / r^2
Electric Field: = k*q / r^2

3. The attempt at a solution

One of the force vectors (call it F1) is pure y. I decomposed the other force vector (call it F2) into it's x and y components.

Vector F1 = < 0, F1 >
Vector F2 = < F2*cos(a), F2*sin(a) >

Last edited: Aug 24, 2007
2. Aug 24, 2007

### rootX

so you know now, the net force acting due to F1 and F2
but the system is in equilibrium so find F3

3. Aug 24, 2007

### mrlucky0

Thanks for the response.

I can find F3:

F3 = F1 + F3
= < F2 cos (a) + F1, sin (a) >

Now, I know can find F1 and F2 using Coulomb's law:

F2 = ( k * 85 uC ) / r2^2
F1 = ( k * 85 uC ) / r1^2

And I also know that the F3 = <0, 0 >

I'm stuck here because I'm not sure what the setup should be.

4. Aug 24, 2007

### rootX

no, you don't know that! It's the net force that =0

F3 != <0,0>

and as k and 85 uC would cancel out, you can find x and y distance of F3

5. Aug 24, 2007

### mrlucky0

I know:

F3 = F1 + F3
= < F2 cos (a) + F1, sin (a) >

and

F2 = ( k * 85 uC ) / r2^2
F1 = ( k * 85 uC ) / r1^2

so for x component:

F2 * cos(a) + F1 = ( k * 85 uC ) / r2^2

y component:

F1 * sin(a) = ( k * 85 uC ) / r1^2

Is this reasonable? Where do I go from here?

6. Aug 24, 2007

### rootX

yea, that's reasonable enough.
you know r1 and r2
and you also know sin(a) and cos(a)

amd so
F3x=that total x-comp
F3y = that total y-comp

and you know it's charge
so you would left with r3x and r3y

7. Aug 24, 2007

### mrlucky0

I'm not following you entirely:

r1, r2, and the angle a is unknown to me. If I am not mistaken, don't these 3 values depend on the position of where I place A? Ultimately I want to the electric field to zero wherever I place A.

8. Aug 24, 2007

### rootX

"are placed at two corners of a square with sides of 0.295 m, as shown. "

A is at one corner of the square.
And reread the question: I think it's like E should be zero at A. So, you are calculating the force acting at point A due to other three charges

so, A is fixed and you cannot move it

Last edited: Aug 24, 2007
9. Aug 24, 2007

### mrlucky0

Oh you're absolutely right. So I'll just determine vector F3 by summing vector F1 and F2:

I have all the ingredients.

r1 = .295
r2 = .42
angle a = 45 degrees

Am I doing the following correctly:

Force F2 = ( 85E-6 C * K ) / ( .42 m)^2
Force F1 = ( 85E-6 C * K ) / ( .295 m)^2

so the vectors:

F2 = < F2 cos(45), F2 sin(45) >
F1 = < 0, F1 >

10. Aug 24, 2007

### rootX

yep seems correct, now find f3-x and f3-y

11. Aug 24, 2007

### mrlucky0

Here's what I'm doing. Using Coulomb's law:

Force F2 = ( 85E-6^2 C * K ) / ( .42 m)^2
= 369 Newtons
Force F1 = ( 85E-6^2 C * K ) / ( .295 m)^2
= 474 Newtons

F1 = < 0, 474 >
F2 = < cos(45)*F2, sin(45)*F2 > = < 261, 261 >

F3 = <261, 735>

These figures don't seem right.

12. Aug 24, 2007

### learningphysics

You're mixing up electric field, and electrical force... For this problem, you need to use electric field...

Electric field strength= kq/r^2

Electric force between to charges = kq1q2/r^2

13. Aug 24, 2007

### rootX

If you have added correctly, and all that, here's what you should do now.

F3[x] = ( 85E-6^2 C * K ) / x^2
F3[y] = ( 85E-6^2 C * K ) / y^2

so find x and y, and it would be the answer.

P.S. you have used the wrong method but nevertheless, you can use it to get the answer

14. Aug 24, 2007

### rootX

"Force F2 = ( 85E-6 C * K ) / ( .42 m)^2
Force F1 = ( 85E-6 C * K ) / ( .295 m)^2"

why did you squared it[85E-6 C * K] in the post after this one?
oops.. I just ignored what you were using, but rather was looking at the way..

It should have been E2 and E1 instead of F1 and F2

sorry,
my mistake

15. Aug 24, 2007

### mrlucky0

I solved those equations and heres what I got:

x = +/- 5
y= +/- 3

The problem states that A is in the top left region relative to the square should I use <-5, 3> ?

16. Aug 24, 2007

### learningphysics

I think a lot of careless mistakes are propping up... first find the field due to those first two charges at A... not the force...

Then the field due to the third charge must cancel the field due to the first two charges... ie: it's the negative of the above field...

Using the magnitude of this field, you can find the distance the third charge is from A... using the direction of the field, you can find the angle from the third charge to A...

then using this angle you can get the x value (distance to the left of A) and y value(distance above A).

17. Aug 24, 2007

### learningphysics

This isn't right... the x component and y component don't work out like that...

18. Aug 24, 2007

### rootX

so F3 = <261, 735>

and thus ( 85E-6^2 C * K ) / x^2 = 261
hence x= 0.4991

and ( 85E-6^2 C * K ) / y^2 = 753
y=0.2938

and I used other method, and got the same answer!

so don't they work?

19. Aug 24, 2007

### mrlucky0

The vector that cancels the first two charges I found to be <-1.3E6, 3.9E6>
The magnitude is 3.7E6.

How do I this is to find the distance the third charge is from A?

20. Aug 24, 2007

### mrlucky0

For the answer, I do think that the x component is negative and the y component is positive since the third large is located somewhere to the left and to the top relative to A. Does that make sense?