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Unformatted text preview: EE 1 - Homework #4 Solutions (1) (20 Points) A parallel-plate capacitor is made using two circular plates of radius a , with the bottom plate on the xy plane, centered at the origin. The top plate is located at z = d , with its center on the z axis. Potential V is on the top plate; the bottom plate is grounded. Dielectric having radially- dependent permittivity fills the region between plates. The permittivity is given by = (1 + r a ) Find: a) V(z) b) E c) Q d) C a. Since varies in the direction normal to ~ E , Laplace’s equation applies, and we write ∇ 2 V = d 2 V dz 2 = 0 = > V ( z ) = C 1 z + C 2 With the given boundary conditions, C 2 = 0, and C 1 = V d . Therefore, V ( z ) = V d z ( V ) b. The electric field is: ~ E =-∇ V = ˆ z- dV dz =- ˆ z V d ( V/m ) c. First we finde the electric flux density: ~ D = ~ E =- ˆ z (1+ r a ) V d ( C/m 2 ). We can now apply Gauss’ Law: Q = Z S ~ D · d ~ S = Z 2 π Z a (1 + r a ) V d rdrdφ Q = 5 πa 2 V 3 d ( C ) d. The capacitance is calculated fromd....
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This note was uploaded on 06/06/2009 for the course EE EE 1 taught by Professor Ozcan during the Spring '09 term at UCLA.
- Spring '09