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 2 years ago
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
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?

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mathavraj
 2 years ago
Best ResponseYou've already chosen the best response.2dw: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..

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

mathavraj
 2 years ago
Best ResponseYou've already chosen the best response.2dw: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

mathavraj
 2 years ago
Best ResponseYou've already chosen the best response.2dw: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

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

VincentLyon.Fr
 2 years ago
Best ResponseYou've already chosen the best response.1In 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.

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