- #31
mrlucky0
- 69
- 1
Yeah. Posted too hastily...Fixed
Where is the Electric Field Equal to Zero?
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- Electric Electric field Field Zero
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SUMMARY
The discussion focuses on determining the position of a third charge that would create an electric field of zero at point A, located at one corner of a square with sides measuring 0.295 m. Two charges of 85.0 µC are placed at two corners of the square. Participants utilize Coulomb's Law and vector decomposition to analyze the forces acting on point A, ultimately concluding that the third charge must be positioned at specific coordinates derived from the electric fields generated by the first two charges. The calculations involve determining the distances and angles necessary to achieve equilibrium at point A.
PREREQUISITES- Coulomb's Law for electric forces: F = (k * q1 * q2) / r^2
- Understanding of electric fields: E = k * q / r^2
- Vector decomposition for force analysis
- Basic trigonometry to resolve forces into components
- Learn about electric field superposition and its applications in electrostatics.
- Study vector addition and decomposition techniques in physics.
- Explore the concept of equilibrium in electrostatic systems.
- Investigate the implications of charge placement on electric field strength and direction.
Students studying electrostatics, physics educators, and anyone interested in understanding electric field interactions and charge placement strategies.
- #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
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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?
-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
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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
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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?
.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
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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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