Question

A positive charge +Q is uniformly distributed along the are of a half-circle of radius R, in the xy-plane with the center at the origin (as shown). Point P is at a height z along the z-axis. (a) Using integration and superposition of the potential due to little dqs, deter- mine the electric potential V at P, as a function of z. (First focus on the setup with a clear plan of attack (good drawing with clear variables). DRAW A BOX AROUND THE INTEGRAL BEFORE INTE- GRATING. Then carry out the integration. SHOW WORK... if you use a tool like Wolfram Alpha, include the code you used to enter the integral to be integrated.) Your answer for V() should be in terms of k, Q, R, and (b) Showing work, take the large z-limit (z >> R) of your function V (2) and report the result, call it Vymptotic(-). (You should obtain a simpler function of z.) On the same graph, plot your original V(2) and kQ/z (with k = 1, Q = 2, R = 3) for the range 0 <? <3R and vertical range set accordingly so that we can see their behaviors. Plot the ratio of functions V(2)/(kQ/-) in the range 0.2R < < < 20 R, with vertical ranges set accordingly. (c) Using integration and superposition of the electric field due to little do's, determine the electric field vector E at P, as a function of . (First focus on the setup with a clear plan of attack good drawing with clear variables). DRAW A BOX AROUND THE INTEGRAL BEFORE INTEGRATING. Then carry out the integration. SHOW WORK... if you use a tool like Wolfram Alpha, include the code you used to enter the integral to be integrated.) Your answer for E should be in terms of k, Q, R, and z. (d) Showing work, take the large 2-limit ( > R) of your vector field E(z) and report the result, call it E ymptotic(). (You should obtain simpler functions of 2.) On the same graph, plot your original --component E,() and kQ/:? (with k = 1. Q = 2, R =3) for the range 0 <3 <3R and vertical range set accordingly so that we can see their behaviors. Plot the ratio of functions E ()/(Q/2') in the range 0.2R<:<20R, with vertical ranges set accordingly.

Answer 1

Charge den sily erdrde = P. TER € 29 I - HR the potential energy at the pointa is a by using corelomb corelomb equation y — г R Cordrdo [v(2) = uraht z 2472429%;h - L1 R +27_z PO CERVER ADV. 2] T I 20K GRX 2 a TUCZ) = 28.5 (1872"- 27 (b) Taking ZIR = 7{1BIILTEL<17 K3 ) Z z = 2.1d + 7617 Hence Vasmplatel2)= 286/21 8K - 2 Esino 7 CCI Ecoso Ecosos Algood for any element of small charge da. the electric field han two component on along z axis and one along x axis. Now the field deur to e and due to 7 te Where oso<h cancel the field along araxis for a perli- cular value of n Hence, we can wrili f2 = 26 orde do 2 =269.3. [do e - KPM2Sacărey - Korz z metodo - PART OR CE Training - 264+ (3 metro dy Now in the limit 774K (84237 06 ( all-z 8, 9, f 0 In al fose yapter 29 - PARTI PESTE Last modified: 9:28 am