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imagreencat
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We were asked to compute for the electric field experience by a point charge P caused by a ringshaped conductor. And my professor just used Q for the computation. However, when we computed for the electric field experienced by P caused by a charged rod, we used the lambda to derive the charge. Shouldn't the ring problem be computed the same way also?
 2 years ago
 2 years ago
imagreencat Group Title
We were asked to compute for the electric field experience by a point charge P caused by a ringshaped conductor. And my professor just used Q for the computation. However, when we computed for the electric field experienced by P caused by a charged rod, we used the lambda to derive the charge. Shouldn't the ring problem be computed the same way also?
 2 years ago
 2 years ago

This Question is Closed

mathavraj Group TitleBest ResponseYou've already chosen the best response.2
dw:1343466065669:dwlambda is linear charge density....in a ring of radius R AND CENTRE O, consider a small element dl on A and B ON the ring..
 2 years ago

mathavraj Group TitleBest ResponseYou've already chosen the best response.2
dw:1343467409133:dw sorry the first sentence is at P due to A
 2 years ago

imagreencat Group TitleBest ResponseYou've already chosen the best response.0
Okay. Thanks!! :)
 2 years ago

mathavraj Group TitleBest ResponseYou've already chosen the best response.2
dw:1343467595714:dw AT p the cos components due to A and B add up while sin components cancel...so if you integrate this ull get the final ans as
 2 years ago

mathavraj Group TitleBest ResponseYou've already chosen the best response.2
dw:1343467750152:dw in integrating use the above egns and use the formula for cos theta....so does this solve ur qn..or u have a diff doubt..pls ask it..i think im confusing u
 2 years ago

mathavraj Group TitleBest ResponseYou've already chosen the best response.2
dw:1343468083471:dwmake this correction..its hard to draw
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
In this problem, the symmetries of all charged points with respect to the axis do not necessitate to break down \(dq\) in \(dq=\lambda dl\), as you would normally do (e.g. straight line). If you did, you would integrate \(\lambda dl\) as \(\lambda \times 2\pi R\), then you would express \(\lambda = \Large \frac{Q}{2\pi R}\) ... and finally find... \(Q\) It is easier to integrate \(dq\) as \(Q\) straight away.
 2 years ago

mathavraj Group TitleBest ResponseYou've already chosen the best response.2
so is my ans correct @VincentLyon.Fr
 2 years ago
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