Simple Nature - Light and Matter

where Q is the total charge of the ring. This result could havebeen derived without calculus, since the distance r is the samefor every point around the ring, i.e., the integrand is a constant.It would also be straightforward to find the field by differentiatingthis expression with respect to z (homework problem 10).Instead, let’s see how to find the field by direct integration. Bysymmetry, the field at the point of interest can have only a componentalong the axis of symmetry, the z axis:E x = 0E y = 0To find the field in the z direction, we integrate the z componentscontributed to the field by each infinitesimal part of the ring.∫E z = dE z∫= | dE| cos θ ,where θ is the angle shown in the figure.∫ k dqE z =r 2 cos θ∫dq= kb 2 + z 2 cos θEverything inside the integral is a constant, so we have∫kE z =b 2 + z 2 cos θ dqkQ=b 2 + z 2 cos θkQ z=b 2 + z 2 rkQz= (b 2 + z 2) 3/2In all the examples presented so far, the charge has been confinedto a one-dimensional line or curve. Although it is possible, forexample, to put charge on a piece of wire, it is more common toencounter practical devices in which the charge is distributed over atwo-dimensional surface, as in the flat metal plates used in Thomson’sexperiments. Mathematically, we can approach this type ofcalculation with the divide-and-conquer technique: slice the surfaceinto lines or curves whose fields we know how to calculate, and thenadd up the contributions to the field from all these slices. In thelimit where the slices are imagined to be infinitesimally thin, wehave an integral.576 Chapter 10 Fields