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EE2011 Lecture 3 - Electrostatics Parrt II

# At conductor surface prof joshua le wei li em research

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Unformatted text preview: s . (at conductor surface). Prof Joshua Le-Wei Li, EM Research Group Prof 20 EE2011: Engineering Electromagnetics Electromagnetics Electric Boundary Conditions • Conductor-Conductor Boundary – The normal component of J has to be continuous across the boundary between two different media under electrostatic conditions. J E1t E2 t , J1t J 2t 1 2 1 , E1n E 2 E2 n J1n 1 s J 2n 2 1 . J1n 21 2 s 1 2 Prof Joshua Le-Wei Li, EM Research Group Prof 1 . 2 EE2011: Engineering Electromagnetics Electromagnetics Capacitance • Concept of Capacitance – Capacitance of a two-conductor capacitor is defined as C = Q/V Q/V (C/V or F) (C/V – Et = 0 while En = s/ on the on surface of the conductor P 1 Q s ˆ n Eds ds S S E ds, and V S V12 E dl P2 P 1 C Q V E ds S , and R P 1 E dl E dl V I P2 E ds so that RC . S P2 Prof Joshua Le-Wei Li, EM Research Group Prof 22 EE2011: Engineering Electromagnetics Electromagnetics Capacitance • Example 8 Question: Derive capacitance C of a parallel-plate capacitor comprised of two parallel plates each of surface area A and separated by a distance d. The capacitor is filled with a dielectric material with permittivity . Also, determine the breakdown voltage if d = 1 cm and the dielectric material is quartz. Solution: Assume the +Q and –Q on the upper and lower plates, respectively. The charge density will be s = Q/A. Hence, E ˆ zE ˆ z d V ˆˆ zE zdz E dl Q V ˆ zQ A d 0 C s Ed , 0 Q Ed A . d Apparently, V = Vbr when E = Eds. As Eds = 30 (MV/m) for quartz, hence the breakdown voltage is Vbr Eds d 30 106 10 Prof Joshua Le-Wei Li, EM Research Group Prof 23 2 3 105 V. EE2011: Engineering Electromagnetics Electromagnetics Capacitance • Example 9 Question: Obtain an expression for the capacitance of the coaxial line shown in the right figure. Solution: Due to the voltage supply, there exists the charge distribution on the inner and outer surfaces of the conductors. The electric field is given by E ˆ r l ˆ r Q . 2r 2 rl Therefore, the potential difference...
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