HW4_solutions

HW4_solutions - EEE241 - Solutions to Homework 4 P. 3-33: A...

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EEE241 - Solutions to Homework 4 P. 3-33: A cylindrical capacitor of length L consists of coaxial conducting surfaces of radii r i and r o . Two dielectric media of different dielectric constants 12 and rr ε fill the space between the conducting surfaces as shown in Fig. 3.42. Determine its capacitance. Assume that the surface of inner conductor has a total charge of Q and the surface of outer conductor has –Q. Note that the charges will not be uniformly distributed over each of the two surfaces because of the two different dielectric materials. It can be easily seen that E and D are radially directed. By considering the boundary conditions between the two dielectric material, it can be seen that E is same in both dielectrics since E is tangential to the boundary. () EEa E r == r 11 22 ror r Da E r E r εε = = r r r r Applying Gauss’s law at io rrr << : () ( ) . or r r D d s E L E L Q π =+= r r ± r r Q Ea rL εε ε = + r The potential difference between the conductors can be calculated as, 12 1 ln ii oo o r r r i r QQ V E dr dr Lr L r πεε ε ⎛⎞ =− = ⎜⎟ ++ ⎝⎠ ∫∫ Thus, the capacitance is given by, 12 ln r o i L Q C V r r +
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P.3-37 A capacitor consists of two concentric spherical shells of radii R i and R o . The space between them
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HW4_solutions - EEE241 - Solutions to Homework 4 P. 3-33: A...

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