- #1
How Do You Calculate Electric Field Intensity?
- Thread starter Ush
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SUMMARY
The discussion focuses on calculating electric field intensity using the formula E = KQ/r², where E represents electric field, Q is the source charge, and r is the distance. Participants emphasize the importance of vector addition when determining the resultant electric field from multiple charges, specifically using the superposition principle. A common issue arises with online grading systems that may penalize for significant figures, leading to discrepancies in answers. The conversation highlights the necessity of careful component analysis in electric field calculations.
PREREQUISITES- Understanding of Coulomb's Law and electric field equations
- Familiarity with vector addition and component analysis
- Knowledge of significant figures in scientific calculations
- Basic proficiency in algebra and trigonometry for resolving components
- Study the superposition principle in electric fields
- Learn about vector addition techniques in physics
- Review Coulomb's Law and its applications in electric field calculations
- Explore common pitfalls in online homework systems regarding significant figures
Students studying physics, particularly those focusing on electromagnetism, as well as educators looking to clarify concepts related to electric fields and vector analysis.
- #2
Onamor
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Well I haven't run through the numbers, but theoretically it seems possible.
The electric field at whatever point is what would act on an imagined point positive charge, so with that in mind:
You have q1 pulling the charge left, and q3 pushing it also left. q2 has no horizontal affect.
Then you've done the nice Pythagoras stuff and found that q1 pulls the charge down, but q2 and q3 push it upwards. Since q2 + q3 have a greater magnitude than q1, they will win.
Pythagoras-ing horizontal and vertical components gives your perfectly depicted arrow, and a magnitude that agrees in sign with this hand waving argument.
Was there something you didn't understand?
The electric field at whatever point is what would act on an imagined point positive charge, so with that in mind:
You have q1 pulling the charge left, and q3 pushing it also left. q2 has no horizontal affect.
Then you've done the nice Pythagoras stuff and found that q1 pulls the charge down, but q2 and q3 push it upwards. Since q2 + q3 have a greater magnitude than q1, they will win.
Pythagoras-ing horizontal and vertical components gives your perfectly depicted arrow, and a magnitude that agrees in sign with this hand waving argument.
Was there something you didn't understand?
Last edited:
- #3
Ush
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Onamor said:
to me, it makes sense. But to the computer marking my answer, it does not =[
- #4
Onamor
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Hmm, do you know what the correct answer is?
I got a different answer to yours with just the superposition equation:
It may just be a computational mistake - when you are calculating the components, you have them on the order nC (x10^-9 in SI units). But the Coulomb constant is
So they should not be of this order.
I got a different answer to yours with just the superposition equation:
It may just be a computational mistake - when you are calculating the components, you have them on the order nC (x10^-9 in SI units). But the Coulomb constant is
So they should not be of this order.
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- #5
Ush
- 97
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I checked over my calculations but i get the same number, I don't know what the correct answer is
I'm not sure how you did your superposition,
but that just looks like the equation for E, - the sum of the electric fields. Electric fields are vectors, you can't add them without breaking them into components- at least I don't know how to yet-
I'm not sure how you did your superposition,
but that just looks like the equation for E, - the sum of the electric fields. Electric fields are vectors, you can't add them without breaking them into components- at least I don't know how to yet-
- #6
Ush
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pleaseeee someone help!?
- #7
Doc Al
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I didn't check your arithmetic, but your method looks correct.
- #8
Ush
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the computer keeps marking it wrong though =/ this seems like a simple question..
- #9
Doc Al
Mentor
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Often the online systems are fussy about significant figures--some of them insist on 3 sig figs no matter what.
- #10
Ush
- 97
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gosh.. thanks o_o it worked!
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