- #1
How Do You Calculate Electric Field Intensity?
- Thread starter Ush
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- Electric Electric field Field
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Homework Help Overview
The discussion revolves around calculating electric field intensity using the formula E = KQ/r², where participants explore the effects of multiple charges on a test charge and the resultant electric field. The context includes theoretical considerations and computational challenges faced in an online homework system.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation
Approaches and Questions Raised
- Participants discuss the theoretical implications of electric field interactions between multiple charges, the application of vector addition for electric fields, and potential computational errors in their calculations. Questions arise regarding the correctness of their answers and the influence of significant figures in online grading systems.
Discussion Status
Some participants express uncertainty about their calculations and the grading criteria of the online system. There is a mix of theoretical exploration and practical troubleshooting, with no clear consensus on the correct answer yet. Guidance has been offered regarding the importance of vector components and significant figures.
Contextual Notes
Participants mention issues with the online grading system, specifically regarding significant figures and the potential for computational mistakes. There is also a reference to the order of magnitude for charge values being a point of confusion.
- #2
Onamor
- 76
- 0
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
- 97
- 0
Onamor said:
to me, it makes sense. But to the computer marking my answer, it does not =[
- #4
Onamor
- 76
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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
- 0
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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