You are here:

Physics/electric fields, charge distributions


uniformly-charged ring
uniformly-charged ring  
Hi, I was just wondering if it would be possible to get some help with tis problem. I don't seem to understand what to do at all. Thanks for any help. A picture is provided to better visualize the problem.

Consider a uniformly-charged ring having a radius of R and a total charge of q_ring.

(a) What is the linear charge density of the ring? Express your answer in terms of R and q_ring
(b) Consider a small element of arc angle delta theta and position theta. What is the charge delta q of this element? Express your answer in terms of R, q_ring, delta theta and theta
(c) What are (i) the electric field delta E, (ii) the electric potential delta V, at P due to delta q? Express your answer in terms of R, q_ring, delta theta, theta and x
(d) What are (i) the net electric field E, (ii) the net electric potential V, at P due to the entire ring? Use symmetry wherever possible, and evaluate your final integrals. Express your answer in terms of R, q_ring and x
(e) Verify that Ex = -dV / dx.

Charge Ring1
Charge Ring1  

Charge Ring 2
Charge Ring 2  
(a) The linear charge density, lambda, is equal to the total charge, Q, divided by the length, the circumference of the loop:
(b)The charge of each little piece of charge dQ will be the charge density, lambda, multiplied by the length of the little piece, dL: dQ=lambda*dL
(c) The little piece of electric field dE at the location of point P caused by the little piece of charge dQ will be:
Since for every piece of charge dQ on one side of the ring there is another piece of charge dQ on the opposite side of the ring. The y components of these two electric field will cancel and so all we need to calculate are the x components. Therefore, we need to multiply by the cosine of the angle:
Integrating both sides of the equation:
Which becomes after integrating:
Since the potential is a scalar the calculation is MUCH simpler! The Potential at point P is just:
Vp=k*Q/r=k*Q/sqrt(x^2+R^2) where Q=lambda*2*pi*R
(D) Taking then derivative of the potential to get the electric field:


All Answers

Answers by Expert:

Ask Experts


James J. Kovalcin


I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

©2017 All rights reserved.