Question

Show missing steps of derivation from equation (22-22) to (22-26) please include explanations. Thank you.

TER 22 IELDS he electric field at an arbitrary point P on the central axis, at distance fromth ter of the disk, as indicated in Fig. 22-15. 22-6 A p pattern of electric field lines around it, but here we restrict our attentio Learning Obje Afher reading this m 22.22 For a charg field (a field du tionship betwe odule but set up a two-dimensional djimensit We could proceed as in the preceding m tegral to include all of the field contributions from the two-dimensionala. tion of charge on the top surface. However, we can save a lot of work wit with a nea rtcut using our earlier work with the field on the central axis of a thin ringea We superimpose a ring on the disk as shown in Fig. 22-15, at an arbitra shortcut ment da. To find its small contribution dE to the electric field at oinel rewrite Eq. 22-16 in terms of the ring's charge da and radius r: R. The ring is so thin that we can treat the charge on it as the particle, a us r Key Ideas feld E, an elect The ring's field points in the positive direction of the z axis adius R and uniform ng shown has radiusr ter of the disk at r 0out to the rim atr R so that we sum all the dE contri sets up a differential tions (by sweeping our arbitrary ring over the entire disk surface). However, thae To find the total field at P. we are going to integrate Eq. 22-22 from the A Point Cha We get dr into the expression by substituting for da in Eq 22-22. Because the is so thin, call its thickness dr. Then its surface area dA is the product of its ence 2 nr and thickness dr. So, in terms of the surface charge density o, we have nt P on its central means we want to integrate with respect to a variable radius r of the ring In the preced charge distrib (22-23) when it is in a What hag After substituting this into Eq. 22-22 and simplifying slightly, we can sum all the dE contributions with (22-24) 480 in which q is field that oth not the fiek where we have pulled the constants (including z) out of the integral. To solve this integral, we cast it in the form Xm dX by setting X- (r), m and dX-(2) dr. For the recast integral we have particle or c The ele field E oppos and so Eq. 22-24 becomes Taking the limits in Eq. 22-25 and rearranging, we find Heasuring Equation American tation of them bec drop that Let us ass If sw (charged disk) (22-26) as the magnitude of the electric field produced by a flat, circular, charged disk at points on its central axis (In carrying out the integration, we assumed that z If we let R → while keeping z finite, the second termin the parentheses in Eq. 22-26 approaches zero, and this equation reduces to positive positive ing plate chambe E =- (infinite sheet). This is the electric field produced by an infinite sheet of uniform charge localed on one side of a nonconductor such as plastic. The electric field lines for a situation are shown in Fig. 22-4 charg We also get Eq. 22-27 if we let z → 0 in Eq. 22-26 while keeping R finite. pe shows that at points very close to the disk, the electric field set up by the the same as if the disk were infinite in extent. By and th

Answer 1