Where is the Electric Field Equal to Zero?

AI Thread Summary
To achieve an electric field of zero at point A, a third charge must be placed above and to the left of A, balancing the fields from the two existing charges at the corners of the square. The forces acting on A are calculated using Coulomb's law, with specific attention to the distances and angles involved. The net electric field at A is determined by summing the contributions from the two existing charges, leading to a required position for the third charge. The calculations involve determining the magnitudes and directions of the electric fields, ensuring that the third charge's field cancels out the net field from the other two. Ultimately, the correct placement of the third charge is crucial for achieving equilibrium at point A.
mrlucky0
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Homework Statement



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

Homework Equations



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

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

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.
 
mrlucky0 said:
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 >

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
 
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?
 
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
 
rootX said:
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

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.
 
mrlucky0 said:
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.

"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
 
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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
mrlucky0 said:
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 >

yep seems correct, now find f3-x and f3-y
 
  • #11
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
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
mrlucky0 said:
F3 = <261, 735>

These figures don't seem right.

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
"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?:eek:
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 :redface:
 
  • #15
rootX said:
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

I solved those equations and here's 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
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
rootX said:
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

This isn't right... the x component and y component don't work out like that...
 
  • #18
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
learningphysics said:
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).

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
rootX said:
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?

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?
 
  • #21
again use Ex=kq/x^2
Ey=kq/y^2

and thus solve for x and y.
 
  • #22
If a charge q is at some point (x, y)... then the magnitude of the field at the (0,0) is:

\frac{kq}{x^2+y^2}

The x-component of the field is \frac{kq}{x^2+y^2} * cos(\theta) = \frac{kq}{(x^2+y^2)}\frac{-x}{\sqrt{x^2+y^2}}=\frac{-kq * x}{(x^2+y^2)^{\frac{3}{2}}

ie: the x-component is NOT -\frac{kq}{x^2}
 
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  • #23
oops, sorry again, I would try them tomorrow morning.
P.S. need a sleep.yea, I got it now! :D
I think I had been thinking wrong through out my last year physics course ><

P.S. somehow I got more than 100% on my static electricity unit test lol
 
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  • #24
rootX said:
oops, sorry again, I would try them tomorrow morning.
P.S. need a sleep.


yea, I got it now! :D
I think I had been thinking wrong through out my last year physics course ><

:wink: :smile:
 
  • #25
mrlucky0 said:
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?

yea, though I can't think any good know, but I guess
it's located "a" distance left,
and less than "a" distance above A.

just don't care about signs much, that makes the job a lot easier. And always try using symmetries and other geometrical tools while solving these problems

P.S. don't sub in values right away, chances of getting a wrong answer become high.
 
  • #26
mrlucky0 said:
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?

Oops... sorry, I didn't notice this till now. Just use kq/r^2 = 3.7E6 to calculate r. Draw a picture...

Are you sure your vector is right? I'm getting this for the field due to the first two charges:

For the first charge
<0, 8.8E6>

this comes from (9E9)(85E-6)/(0.295^2)

For the second charge
<-3.1E6,3.1E6>

this comes from (9E9)(85E-6)/(0.295^2 + 0.295^2)cos(45)

So the total I'm getting is:
<-3.1E6, 11.9E6> so with a magnitude of 12.3E6
 
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  • #27
learningphysics said:
Oops... sorry, I didn't notice this till now. Just use kq/r^2 = 3.7E6 to calculate r. Draw a picture...

Are you sure your vector is right? I'm getting this for the field due to the first two charges:

For the first charge
<0, 8.8E6>

this comes from (9E9)(85E-6)/(0.295^2)

For the second charge
<-3.1E6,3.1E6>

this comes from (9E9)(85E-6)/(0.295^2 + 0.295^2)cos(45)

So the total I'm getting is:
<-3.1E6, 11.9E6> so with a magnitude of 12.3E6

This is how I proceeded. I decompose E2. The .42 ( the r from the second charge to A) is just the hypotenuse of a right triangle: sqrt(.295^2 + .295^2) = .42

I am saying that the angle 135 degrees is the angle of E2 relative to the positive x-axis.

E1 = <0, 8.8E6 >

(same thing you did)

E2 = (9E9)(85E-6)/.42^2 * <cos(135), sin(135) >
= 4.3E7 * < cos(135), sin (135) >
= < -3E7, 3E7 >

Enet = E1 + E2 = < -3E7, 3.9E7 >
||Enet|| = 4.7E7

So it seems our only disagreement is E2. Where am I going wrong?
 
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  • #28
Check your exponent when you calculate E2... I think it should be 4.3E6.
 
  • #29
learningphysics said:
Check your exponent when you calculate E2... I think it should be 4.3E6.

Whoops. Ok:

E1 = <0, 8.8E6 >

E2 = (9E9)(85E-6)/.42^2 * <cos(135), sin(135) >
= 4.3E6 * < cos(135), sin (135) >
= < -3E6, 3E6 >

Enet = E1 + E2 = < -3E6, 1.2E7 >
||Enet|| = 1.2E7

Looks good now right? Could you explain where I go from here?
 
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  • #30
mrlucky0 said:
Whoops. Ok:

E1 = <0, 8.8E6 >

E2 = (9E9)(85E-6)/.42^2 * <cos(135), sin(135) >
= 4.3E6 * < cos(135), sin (135) >
= < -3E6, 3E6 >

Enet = E1 + E2 = < -3E7, 3.9E7 >
||Enet|| = 4.7E7

Looks good now right? Could you explain where I go from here?

You should fix Enet.
 
  • #31
Yeah. Posted too hastily...Fixed
 
  • #32
Cool. So you want the field created by the 3rd charge to cancel this one... so it has the same magnitude and opposite direction... I recommend first calculating r using:

kq/r^2 = 1.2E7
 
  • #33
The field created by the third charge is then

-Enet = < 3E6, -1.2E7 >

(9E9)(85E-6)/r^2 = 1.2E7 ==> r = .25

Looks good?
 
  • #34
mrlucky0 said:
The field created by the third charge is then

-Enet = < 3E6, -1.2E7 >

(9E9)(85E-6)/r^2 = 1.2E7 ==> r = .25

Looks good?

Yeah... looks good. I'm a little worried about the number of decimal places we're keeping... I think more would be better, but no big deal...

Do you have an idea about where the third charge should be (the angle)? you can get the angle from the field...
 
  • #35
learningphysics said:
Yeah... looks good. I'm a little worried about the number of decimal places we're keeping... I think more would be better, but no big deal...

Do you have an idea about where the third charge should be (the angle)? you can get the angle from the field...

Regarding decimal places, I will go through the problem again with more accuracy no worries at this point.

How about getting the angle using inverse tangent:

I know < 3E6, -1.2E7 >.

tan(a) = (1.27E7/ 3E6) ==> a = 76 degrees
 
  • #36
mrlucky0 said:
Regarding decimal places, I will go through the problem again with more accuracy no worries at this point.

How about getting the angle using inverse tangent:

I know < 3E6, -1.2E7 >.

tan(a) = (1.27E7/ 3E6) ==> a = 76 degrees

Yeah, so now you should be able to get x and y from 76 and r=0.25
 
  • #37
I forgot. 76 degrees is relative to the -x axis so 104 degrees is relative to the +x axis:

.25 < cos(104), sin(104) >
= <-6E-2, 2.4E-1>

The signs of the coordinate make sense to me. Success?
 
  • #38
mrlucky0 said:
I forgot. 76 degrees is relative to the -x axis so 104 degrees is relative to the +x axis:

.25 < cos(104), sin(104) >
= <-6E-2, 2.4E-1>

Cool! Yeah, I tend to just use the acute angles, and the sketch to keep track of the signs (positive or negative x, y etc...)... But how you did it is exactly right.
 
  • #39
mrlucky0 said:
I forgot. 76 degrees is relative to the -x axis so 104 degrees is relative to the +x axis:

.25 < cos(104), sin(104) >
= <-6E-2, 2.4E-1>

The signs of the coordinate make sense to me. Success?

Yup. :cool:
 
  • #40
Wow. Thank you so much. I just joined this forum today and I'm so glad I did. This was a positive experience. You are of great help.
 
  • #41
mrlucky0 said:
Wow. Thank you so much. I just joined this forum today and I'm so glad I did. This was a positive experience. You are of great help.

You're welcome! :smile: Yeah, it's a great forum.
 

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