Fluxes and Fields


A disc (radius R, surface charge Sigma), and a point charge Q are separated by distance x...

Do:
1)Find field at the location of the charge due to the disc.
2)Flux due to the charge's field thru the disc.
3)Learn them both...you already know (1)!

9 comments:

  1. Sambhav GuptaMay 18, 2010 at 11:38 PM

    This comment has been removed by the author.

    ReplyDelete
  2. Sambhav GuptaMay 19, 2010 at 12:04 AM

    hey do we have to use the gauss law or the definition?

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  3. Sambhav GuptaMay 19, 2010 at 12:08 AM

    Q/(sqrt[8]e)

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  4. Sambhav GuptaMay 19, 2010 at 2:04 AM

    1) (sigma/2e)(1-cosQ)
    2) (q/4e)(1-cosQ)

    Q=angle made by the axis with line joining the particle to the rim of the disc

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  5. AnujMay 19, 2010 at 7:29 PM

    hi yaar...you got most of it right...the flux actually is q/2e(1-CosQ)..messed up some integration???

    and it seems like you people have not really done some electrostats...lets leave it!

    ReplyDelete
  6. Sambhav GuptaMay 19, 2010 at 9:14 PM

    yea i did it...
    nice question btw...was wondering if we could could do that by gauss' law

    only 1 batch got this chap completed yet...and tha too today
    will be over in all by a month

    ReplyDelete
  7. AnujMay 19, 2010 at 10:08 PM

    for gauss law to be applied, you need some symmetry...i can't find any here, and probably no one has...

    ReplyDelete
  8. Mridul GuptaMay 20, 2010 at 10:26 PM

    IS THE ANSWER -sigma/2e * x/R

    ReplyDelete
  9. AnujMay 21, 2010 at 4:43 PM

    hi Mridul..the answer is infact sigma/2e(1-CosQ)

    Q is as described by Sambhav in his post.

    ReplyDelete

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