yrelhan4
  • yrelhan4
flux problem?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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yrelhan4
  • yrelhan4
|dw:1356550535697:dw| if a charge q is placed at the point M in the diagram, the flux caused through the sphere?
anonymous
  • anonymous
is the charge placed on the sphere?
yrelhan4
  • yrelhan4
point m is halfway between o and o1. o and o1 are on peripherry of ring and sphere respectively.. they are like in a strt line.. the sphere and the ring are symmetrical and o and o1 lie on the diameter..

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yrelhan4
  • yrelhan4
answer is \[4q/3\]
yrelhan4
  • yrelhan4
3epsilon*
yrelhan4
  • yrelhan4
the flux in the sphere.. i'm sorry.. wanted to write something else. just messed up..
anonymous
  • anonymous
is it a metal sphere? because if it is, the flux through it is zero...
yrelhan4
  • yrelhan4
i dont think it is.. the answer is not zero.. so its not..
anonymous
  • anonymous
but still, the charge is outside the sphere, isn't it? so q(enclosed)=0. And therefore, the flux though the sphere is zero.
yrelhan4
  • yrelhan4
no it isnt outside the sphere.. its at M..
anonymous
  • anonymous
yeah....i'm not getting the picture here...
anonymous
  • anonymous
|dw:1356553072695:dw| if you look at the arrangement from the side, is this how it'd look?
yrelhan4
  • yrelhan4
no.. they are in the same plane.. intersecting at two points..
yrelhan4
  • yrelhan4
should i post a pic from the book?
anonymous
  • anonymous
Ahhh now i see it, thanks.
yrelhan4
  • yrelhan4
just reply if you get it.. i'm going off for now.. i have a similar problem if that would help you.. suppose if the charge is placed uniformly on the ring.. then the part with the sphere is only q/3.. so the flux there wuld be q/3epsilon.. i dont think its of any help.. just in case..
anonymous
  • anonymous
i think it'd be great if you could post the pic..
yrelhan4
  • yrelhan4
just a minute..
yrelhan4
  • yrelhan4
yrelhan4
  • yrelhan4
try it, and tell me you get it..
anonymous
  • anonymous
i think the question is flawed. Here's why: the only way you can end up with an answer of \(\Large \frac{4q}{3\epsilon_0}\) is if there's a charge of +q/3 on the ring. [That makes q(enclosed) = q + (q/3) = 4q/3 and so the flux turns out to be \(\large \frac{4q}{3\epsilon_0}\).] But that's not what happens, When you place that positive charge (+q) there, the part of the ring on the inside of the sphere becomes negatively charged and the part outside becomes positively charged (Of course, here I'm assuming that the ring and the sphere are not touching each other). So, the net charge inside the sphere is \( \text{q-(something)}\), which is obviously less than \(\Large \frac{4q}{3}\).
yrelhan4
  • yrelhan4
hmm.. i seem to be getting your point.. thank you!!

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