This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Problem 1
a) Boundary conditions include continuity of potential and electrical ﬁeld
(derivative of potential) at x = 0, and continuity of electrical ﬁeld at x = −xp
and x = xn , namely
1. Vp (0) = Vn (0)
2. ∂
∂x Vp (0) 3. ∂
∂x Vp (−xp ) 4. ∂
∂x Vn (xn ) = ∂
∂x Vn (0) =0 =0 We also have the constraint that Vn (xn ) − Vp (−xp ) = Vbi . This gives rise to the
dependence of parameters xn , xp and Vbi which is asked to be found in part d).
b) The charge density within each region is constant. By solving the Poisson’s
equation, we have that the potential within each region is a quadratic function
with respect to x.
eNA 2
x + Ax + B
Vp (x) =
2Ks ε0
eND 2
Vn (x) = −
x + Cx + D
2Ks ε0
By boundary condition 1, we have B = D. We can always assume that the
potential at x = 0 is zero so that B = D = 0. By boundary condition 3&4, we
have
eNA
xp
A=
Ks ε0
eND
xn
C=
K s ε0
By boundary condition 2, we have A = C , so that
ND x n = NA x p
c)
∂
eNA
Vp (x) = −
( x + xp )
∂x
Ks ε0
eND
∂
(x − xn )
En (x) = − Vn (x) =
∂x
Ks ε0 Ep (x) = − d) e
(ND x2 + NA x2 )
n
p
2Ks ε0
Combining with ND xn = NA xp we can easily obtain
Vbi = Vn (xn ) − Vp (−xp ) = xp + xn = 2Ks ε0 Vbi NA + ND
e
NA ND 1 1/2 ...
View
Full
Document
This note was uploaded on 02/02/2012 for the course ELECTRICAL ENEE380 taught by Professor D.romeo during the Fall '11 term at Maryland.
 Fall '11
 D.Romeo
 Electromagnet

Click to edit the document details