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Question:
An Instructor’s Solutions Manual to Accompany Fundamentals of Chemical
Engineering Thermodynamics [EXP-5137] (https://holooly.com/sources/an-instructorssolutions-manual-to-accompany-fundamentals-of-chemical-engineeringthermodynamics-exp-5137/)
A lucite sheet (ϵr
​
= 3.2) is introduced perpendicularly in a uniform electric field E o =
​
ax E o in free space. Determine E i , Di and P i inside the lucite.
​
​
​
​
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Verified Solution 
We assume that the introduction of the lucite sheet doe not disturb the original uniform
electric field E o The situation is depicted in Fig. 3-17. Since
​
the interfaces are perpendicular to the electric field , only the normal field components
need be considered. No free charges exist. Boundary condition Eq. (3-77) at the left
interface gives
Di = ax Di = ax Do ,
​
​
​
​
​
or
Di = ax ϵo E o .
​
​
​
​
There i no change in electric flux density across the interface. The electric field intensity
inside the lucite sheet is
E i = 1ϵ Di =
​
​
1
ϵo ϵr
​
​
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Eo
Di = ax 3.2
​
​
​
​
​
​
Hence the effect of the lucite sheet is to reduce electric intensity. The polarization vector
is zero outside the Iucite sheet (P o
P i = Di − ϵo E i = ax (1 −
​
​
​
​
​
= ax 0.6875ϵ0 E o
​
​
​
= 0). Inside the sheet,
1
3.2 ) ϵo E o
​
​
​
(C/m2 )
​
Clearly, a similar application of the boundary condition Eq. (3-77) on the right interface
will yield the original E o and D o in the free space on the right of the lucite sheet.
​
​
Does the solution of this problem change if the original electric field is not uniform; that
is, if E o
​
= ax E(y)?
D1n = D2n
​
​
​
(3 − 77)
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