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  1. #1

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    Default Physics Homework -

    Sorry to bother you guys, but I have this homework question that I'm stumped on:

    "You're 1.5m from a charge distribution whose size is much less than 1m. You measure an electric field strength of 282N/C. You move to a distance of 2.0m, and the field strength becomes 119N/C. What is the net charge of the distribution?"

    I was trying to find out a way to calculate the charge, but I'm not sure where to go with this. All I understand is that the electric field decreases with distance?? Any help would be greatly appreciated!
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  2. #2

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    The answer is C. Statistically C is the correct answer 80% of the time.

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    I hate physics....if you don't know the answer damn it...CHEAT!!!

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    42 -Its the answer to everything!
    There is no genuine justice in any scheme of feeding and coddling the loafer whose only ponderable energies are devoted wholly to reproduction. Nine-tenths of the rights he bellows for are really privileges and he does nothing to deserve them. We not only acquired a vast population of morons, we have inculcated all morons, old or young, with the doctrine that the decent and industrious people of the country are bound to support them for all time.-Menkin

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    i get 807.44n/c

    A 42% increase ever .5m you move closer.

    But I could be wrong
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    282 @.1.5m
    400.44 @1m
    568.62 @.5m
    807.44@0m
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  7. #7

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    Quote Originally Posted by jakelm View Post
    282 @.1.5m
    400.44 @1m
    568.62 @.5m
    807.44@0m
    Looks more like Russ's booze level on a Friday night.:p

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    Quote Originally Posted by jakelm View Post
    i get 807.44n/c

    A 42% increase ever .5m you move closer.

    But I could be wrong
    How did you get that? I'm less interested in the answer, more interested in understanding the problem.

    As I said before I was trying to calculate the charge, but there must be an easier way...
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    Quote Originally Posted by appadv View Post
    How did you get that? I'm less interested in the answer, more interested in understanding the problem.

    As I said before I was trying to calculate the charge, but there must be an easier way...
    Ok . You have the measurement between 2.0m (119) and 1.5m (282). 119 devided by 282 is .42

    42% is the increase every .5m you get closer.

    Now go from 1.5m to 1.0 and add 42%
    Now go from 1.0m to .5. and add 42%
    now go from .5m to 0m and again add 42%

    a charge is constant, it does not change. so an increase between 2.0m and 1.5m is the same as the increase between .5m and 0m. So the % of increase or decrease never changes.
    Last edited by jakelm; 01-29-2008 at 05:20 PM.
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    Quote Originally Posted by BaggedLancer View Post
    The answer is C. Statistically C is the correct answer 80% of the time.
    They've done studies, you know. 60% of the time it works, every time.

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    Quote Originally Posted by Demiurge View Post
    They've done studies, you know. 60% of the time it works, every time.
    60% is an F.

    But 100% of 60% of the time is pretty good.
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    Quote Originally Posted by jakelm View Post
    Ok . You have the measurement between 2.0m (119) and 1.5m (282). 119 devided by 282 is .42

    42% is the increase every .5m you get closer.

    Now go from 1.5m to 1.0 and add 42%
    Now go from 1.0m to .5. and add 42%
    now go from .5m to 0m and again add 42%

    a charge is constant, it does not change. so an increase between 2.0m and 1.5m is the same as the increase between .5m and 0m. So the % of increase or decrease never changes.
    Okay, cool! I was getting nowhere when trying to calculate the charge, but then after you explained I realized the charge is constant and you can solve it your way.

    Thanks!
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    Quote Originally Posted by jakelm View Post
    60% is an F.

    But 100% of 60% of the time is pretty good.

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    Default

    Or I could be completely wrong and it could be something like this.


    http://www.ux1.eiu.edu/~cfadd/1360/25ElPot/Continu.html
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  15. #15

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    I think I am competely wrong at this equation, I must look like an idiot...lol
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    Nah, I don't get it either...
    Last edited by Serendipity; 01-29-2008 at 08:31 PM.
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  17. #17

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    So what is it? Did you figure it out? What equation are you using?
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  18. #18

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    Quote Originally Posted by jakelm View Post
    So what is it? Did you figure it out? What equation are you using?
    I thought I was close, but you can say that I didn't figure it out yet.
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  19. #19

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    I understand.

    Is there something missing from your question? A charge distribution less than 1m? Thats it? Is it an atom? What formula are you using? I wish I could help.. Good luck
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    This sounds like a Physics 208 problem (electromagnetism, charges, electricity, etc). Couldn't you use point charge formulas? I threw away all my notes from last semester so I can't really help you too much :(
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    Inside the field is always zero.

    Maybe this..

    sigma=-qd/[r^2+_d^2]^{3/2}

    Dont ask me....I read somewhere about using an image charge..
    Last edited by jakelm; 01-29-2008 at 09:19 PM.
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  22. #22

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  23. #23

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    Quote Originally Posted by nms View Post
    This sounds like a Physics 208 problem (electromagnetism, charges, electricity, etc). Couldn't you use point charge formulas? I threw away all my notes from last semester so I can't really help you too much :(
    I'm not sure if I could use point charge formulas in this case???

    Anyways, the question in the first post is the entire question.
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