Question

9. (a) Consider a uniformly charged, thin-walled, right circular s cylindrical shell having total charge Q. radius R, and length . Determine the electric field at a point a distance d from the right side of the cylinder as shown in Figure P23.9. Suggestion: Use the result of Example 23.2 and treat the cylinder as a col- lection of ring charges. (b) What If? Consider now a solid cyl- inder with the same dimensions and carrying the same charge, uniformly distributed through its volume. Use the result of Example 23.3 to find the field it creates at the same point,

Answer 1

Right circular cylin decal shell Total a radiis ? R, C lengths l. ewall element ring at distance & freom porn lets take a small element rino thickness da 10) Charge on this suing element = Q x 2rxdx 200 - Q dve i due to tlus small element at point p at distance ofx. using Ê due to ring at distance & dE - KRO Kx Qd x TV 7 R233)2 37, elore y R 2012 Jutegrating the Hecbeic field produced due to all euchi Continous dx sing element at pourt P Integrating limits of dx from d ate . " + KxQax I elx²+R2 32 dre kar xdx E-0= 7 178 R2y3/2 let X²+R²=t, differentiating axdxz &dt adx = dt/2

Putting t = R² + x² e unity enter - in right druction Novothe isleid cylinder y Charger dame Lureusou. Lets take a small element of dise of thickness dx at distance & from pout P. Volume durity, f = charge (6) charge on this small deze element - = Qdre Ē due to this small disc elemont at distance & dĒ = 20Koodil Llor

2A d l (x²+ R²)/2 o't adre, de = 20e double Integrating limits of x from d d +1) dte d- 2KQ er bara svona = PKA (1+2+4) - 2 ) - 2KQ 2 , axume + R2 - axdxdt. - I I 12 I - -2+1 KQ / dt 1 KO lat R I +)2 R? Alty2 (+)2 pee (1/2+1 ] R t x²71²" In dkg (20+23% Ree

E (8 +27°+23)3 + P R 1 44 22% N0 2K2 For courtya [va parte) R towards right direction