Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

|dw:1343466065669:dw|lambda is linear charge 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!! :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1343467595714:dw| AT p the cos components due to A and B add up while sin components 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 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

Not the answer you are looking for?

Search for more explanations.

Ask your own question