sp02em_sol

# sp02em_sol - E & M Proficiency Exam, Spring 2002 Useful...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: E & M Proficiency Exam, Spring 2002 Useful constants: • e = 1 . 60 × 10- 19 C • m e = 0 . 511 MeV c 2 • c = 3 . 00 × 10 8 m s • k = 1 4 π = 8 . 99 × 10 9 Nm 2 C 2 • = 8 . 85 × 10- 12 C 2 Nm 2 1. A parallel-plate capacitor is constructed using a dielectric whose constant varies with position. The plates have area A . The bottom plate is at y = 0 and the top plate is at y = y . The dielectric constant is given as a function of y according to κ = 1 + 3 y y . (a) What is the capacitance? Capacitance is defined by C = Q V , (1) where Q and V are free charge and potential of the capacitor respectively. The electric field intensity points from the plate with the positive charge to the plate with the negative charge. +- x y E y=0 y=y The potential is given by V =- Z y = y y =0 d~x · ~ E =- Z y = y y =0 dy E. (2) Now, we determine the electric field intensity ~ E . When an dielectric is placed in an external electric field, the external electric field polarizes the dielectric. Therefore there will be bound charges on the surface of the dielectric. In this kind of problems it is sometimes easier to work with the electric displacement vector ~ D instead of ~ E . The reason is that the divergence of ~ D is given by ∇ · ~ D = σ f , (3) where σ f is the free charge density. Because free charges only exist on the surface of the plates. The electric field will be obtained by ~ E = ~ D = ~ D κ , (4) where κ is the relative permitivity of the dielectric material. Let the total surface charge density on the upper plates is σ f then by applying Gauss’s theorem we get on a small box ~ D =- ˆ y σ f . (5) And the electric field will be ~ E =- ˆ y σ f 1 κ ( y ) (6) Substitute Eq.6 in Eq.2 and perform the integration. We get V = σ f Z y = y y =0 dy 1 κ ( y ) = σ f Z y = y y =0 dy 1 (1 + 3 y y ) . = σ f y 3 ln 1 + 3 y y ! = σ f y 3 ln 4 (7) The total charge is given by Q = σ f A, A is the area of the plate (8) Substitute Eqs.7 and 8 in Eq.1. We obtain C = 3 A y ln 4 (9) (b) Find the ratio σ b /σ f between the bound and free area charge densities on the the surfaces of the dielectric....
View Full Document

## This note was uploaded on 03/11/2012 for the course PHY 3900 taught by Professor Staff during the Fall '1 term at FSU.

### Page1 / 7

sp02em_sol - E & M Proficiency Exam, Spring 2002 Useful...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online