- #31
mrlucky0
- 69
- 1
Yeah. Posted too hastily...Fixed
Where is the Electric Field Equal to Zero?
- Thread starter mrlucky0
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- Electric Electric field Field Zero
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Homework Help Overview
The problem involves determining the position of a third charged object such that the electric field at a specific point, labeled A, is zero. The context is based on two existing charges placed at corners of a square, and the task is to find the location of the third charge relative to point A.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants discuss the decomposition of force vectors and the application of Coulomb's law to find forces acting on point A. There are attempts to clarify the relationship between the forces and the electric field, with some participants questioning the setup and assumptions regarding the distances and angles involved.
Discussion Status
There is ongoing exploration of the relationships between the forces and the electric field, with some participants suggesting corrections to the approach. Multiple interpretations of the problem setup are being considered, particularly regarding the calculations of distances and angles. Some guidance has been offered about focusing on the electric field rather than forces.
Contextual Notes
Participants note that the position of point A is fixed, and the calculations depend on the specific configuration of the charges. There is also mention of potential confusion between electric force and electric field calculations, which may affect the understanding of the problem.
- #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
kq/r^2 = 1.2E7
- #33
mrlucky0
- 69
- 1
The field created by the third charge is then
-Enet = < 3E6, -1.2E7 >
(9E9)(85E-6)/r^2 = 1.2E7 ==> r = .25
Looks good?
-Enet = < 3E6, -1.2E7 >
(9E9)(85E-6)/r^2 = 1.2E7 ==> r = .25
Looks good?
- #34
learningphysics
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mrlucky0 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...
- #35
mrlucky0
- 69
- 1
learningphysics 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
- #36
learningphysics
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mrlucky0 said:
Yeah, so now you should be able to get x and y from 76 and r=0.25
- #37
mrlucky0
- 69
- 1
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?
.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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mrlucky0 said:
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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mrlucky0 said:
Yup.
- #40
mrlucky0
- 69
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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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mrlucky0 said:
You're welcome!
Yeah, it's a great forum.Similar threads
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