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Where is the Electric Field Equal to Zero?

  • Thread starter mrlucky0
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  • #26
learningphysics
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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
 
Last edited:
  • #27
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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?
 
Last edited:
  • #28
learningphysics
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Check your exponent when you calculate E2... I think it should be 4.3E6.
 
  • #29
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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?
 
Last edited:
  • #30
learningphysics
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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
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Yeah. Posted too hastily...Fixed
 
  • #32
learningphysics
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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
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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
learningphysics
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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
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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
learningphysics
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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
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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
learningphysics
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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
learningphysics
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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
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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
learningphysics
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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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