Problem Find the electric potential and the electricfield at a point P located on the axis of a uniformlycharged ring of radius a and total charge Q. The planeof the ring is chosen perpendicular to the x axis (Fig.25.16).daVa? + a²Strategy Figure 25.16 helps us to visualize the sourceof the potential and conceptualize what the potentialmight look like. We expect the potential to besymmetric around the x axis and to decrease forincreasing values of x. We categorize this problem asone involving a continuous distribution of charge onthe ring rather than a collection of individual charges.Figure 25.16 A uniformly charged ring of radius a,whose plane is perpendicular to the x axis. Allelements dq of the ring are at the same distanceaway from any point P on the x axis.SolutionTo analyze the problem, let us take P to be at a distance x from the center of the ring as in Figure 25.16. Thecharge element dq is at a distance equal to r = vx2 + a? from point P. Hence, using Equation 25.20, we canexpress V as follows.dqV = ke= kex2+a?In this case, each element dq is at the same distance from P. The term Vr2 + a? can therefore be removedfrom the integral, and V reduces to what is given below. (Use the following as necessary: ke, Q, a, and x.)keV =dq :=+a² Let us now address the electric field. The only variable in the expression for V is x. From the symmetry, wesee that along the x axis E can have only an x component. We can therefore use Equation 25.16 to find themagnitude of the electric field at P. (Use the following as necessary: ke, Q, a, and x.)d–k„Q– (x² + a²)-1/2 = -k,Q(2² + a²) -3/² (2.x)drE,To finalize, note that V decreases as x increases, as we expected from our mental representation. If the pointPis very far from the ring (x >> a), then a in the denominator of the expression for V can be ignored and V ×kQ/x. This expression is just the one you would expect for a point charge. At large values of x, therefore, thecharge distribution appears to be a point charge of magnitude Q as you should expect. Also notice that thisresult for the electric field agrees with that obtained by direct integration.Exercise 25.5Hints: Getting Started | I'm StuckWhat is the electric potential at the center of the uniformly charged ring? (Use thefollowing as necessary: ke, Q, and a.)V =

Answered: Image /qna-images/answer/467bf8df-e309-46b9-b005-ec98088bac6b.jpg

Problem Find the electric potential and the electric field at a point P located on the axis of a uniformly charged ring of radius a and total charge Q. The plane of the ring is chosen perpendicular to the x axis (Fig. | 25.16).

Problem Find the electric potential and the electric field at a point P located on the axis of a uniformly charged ring of radius a and total charge Q. The plane of the ring is chosen perpendicular to the x axis (Fig. | 25.16).

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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