Engineering Electromagnetics, W. H. Hayt and J. A. Buck, McGraw Hill, 2001 (6th edition) 7장 족보
- 최초 등록일
- 2010.09.26
- 최종 저작일
- 2010.09
- 24페이지/ 어도비 PDF
- 가격 1,500원
소개글
Engineering Electromagnetics, W. H. Hayt and J. A. Buck, McGraw Hill, 2001 (6th edition) 연습문제 풀이 7장 풀이.
목차
없음
본문내용
7.9. The functions V1(ρ, φ, z) and V2(ρ, φ, z) both satisfy Laplace’s equation in the region a < ρ < b,
0 ≤ φ < 2π, −L < z < L; each is zero on the surfaces ρ = b for −L < z < L; z = −L for
a < ρ < b; and z = L for a < ρ < b; and each is 100 V on the surface ρ = a for −L < z < L.
a) In the region specified above, is Laplace’s equation satisfied by the functions V1 + V2,
V1 − V2, V1 + 3, and V1V2? Yes for the first three, since Laplace’s equation is linear. No
for V1V2.
b) On the boundary surfaces specified, are the potential values given above obtained from
the functions V1+V2, V1−V2, V1+3, and V1V2? At the 100 V surface (ρ = a), No for all.
At the 0 V surfaces, yes, except for V1 + 3.
c) Are the functions V1 + V2, V1 − V2, V1 + 3, and V1V2 identical with V1? Only V2 is,
since it is given as satisfying all the boundary conditions that V1 does. Therefore, by the
uniqueness theorem, V2 = V1. The others, not satisfying the boundary conditions, are
not the same as V1.
5
7.10. Consider the parallel-plate capacitor of Problem 7.6, but this time the charged dielectric exists
only between z = 0 and z = b, where b < d. Free space fills the region b < z < d. Both
plates are at ground potential. No surface charge exists at z = b, so that both V and D are
continuous there. By solving Laplace’s and Poisson’s equations, find:
a) V (z) for 0 < z < d: In Region 1 (z < b), we solve Poisson’s equation, assuming z variation
only:
참고 자료
없음