We were asked to compute for the electric field experience by a point charge P caused by a ring-shaped 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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We were asked to compute for the electric field experience by a point charge P caused by a ring-shaped 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?

Physics
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|dw:1343466065669:dw|lambda 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..
|dw:1343467409133:dw| sorry the first sentence is at P due to A
Okay. Thanks!! :)

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|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
|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
|dw:1343468083471:dw|make this correction..its hard to draw
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.
so is my ans correct @Vincent-Lyon.Fr

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