Question

A 30 whc 45° moldo Problem 4 Part A A coaxial cable consists of two infinitelv long conducting coaxial cylinders as shown. The inner evlinder is solid with a radius of n and is positively charged with a charge per unit length of A. The outer evlinder is hollow with an inner radius of ra and an outer radius of ra. It is hegatively charged but. with the same magnitude charge per unit length. (a) Make a sketch of the cross-section of the cable in vour answer booklet. On your sketch show: . The location of the charges. The direction (inwards, outwards or zero) of the electric field in cach of the four regions. (r<n.nr<r, T<r<s.r>rs). 7erd Make sure you justify your choices for the electric field directions on your diagram. 700 (b) Use Gauss's law to calculate an expression for the electric field as a function of radius in any region where the field is non-zero. (c) Suppose the charge density of the outer cylinder were doubled (so the charge per unit length of the negative charge on the outer cylinder is 2A while the charge on the inner cylinder is un- changed). Discuss how the electric field distribution would change (you do not need to perform any calculations, but a diagram would aid your explanation). Side Cross-section Part B (Advanced) Consider now the same shaped coaxial cable as in Part A. However non-conducting material with the charge uniformly distributed (by volume) throughout the material. The inner cylinder is, as before, positively charged with a charge per unit length of A. The outer cylinder is again negatively charged with the same magnitude charge per unit length. However, as the inner and outer conductors have different cross-sectional areas, they will have different charge densities (charge per unit volume). Specifically, the magnitudes of these charge densities will be cylinders are now made of a Pinner Poster= Will the electric field in each of the four regions for this configuration differ from your answers in Sections (a) and (b) of Part A? If so, determine the equation for the electric field. If not, explain why not. Note: for a cylinder of radius r, and length l: Surface area 2arl. Volume = rr. Nert question over... Page 5 of 12

Answer 1

oorductor, The only the urface. Ca.). Sinee charge Can't exist ingide optionfor te charges ji to vemain on Since insk hence E zero when T<Y, and Y< T Hgo 8in ce pointa outward for inward for a conductor eleetric fied jg zero charge and the ekctric field will be inward for 2<r<T3 4Ye chage hence and C Lecolaulate the electric field for hat the ek ctric feld cue to outer from Cwe notice cylinder the inner cylindev-. will be zero the only contibution will Come using the gauaa law. now fE.d Ex 2TT 27 CC) The electric field jngide will remain the ame while outside the magnitude uill get doubled