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yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
dw:1356550535697:dw if a charge q is placed at the point M in the diagram, the flux caused through the sphere?
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.1
is the charge placed on the sphere?
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
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..
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
answer is \[4q/3\]
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
3epsilon*
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
the flux in the sphere.. i'm sorry.. wanted to write something else. just messed up..
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.1
is it a metal sphere? because if it is, the flux through it is zero...
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
i dont think it is.. the answer is not zero.. so its not..
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.1
but still, the charge is outside the sphere, isn't it? so q(enclosed)=0. And therefore, the flux though the sphere is zero.
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
no it isnt outside the sphere.. its at M..
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.1
yeah....i'm not getting the picture here...
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.1
dw:1356553072695:dw if you look at the arrangement from the side, is this how it'd look?
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
no.. they are in the same plane.. intersecting at two points..
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
should i post a pic from the book?
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.1
Ahhh now i see it, thanks.
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
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..
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.1
i think it'd be great if you could post the pic..
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
just a minute..
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
try it, and tell me you get it..
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.1
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 \(\Large \frac{4q}{3}\).
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.0
hmm.. i seem to be getting your point.. thank you!!
 one year ago
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