The figure shows a parallel-plate capacitor with a plate area A = 2.52 cm2 and plate separation d = 7.95 mm. The bottom half of the gap is filled with material of dielectric constant κ1 = 13.7; the top half is filled with material of dielectric constant κ2 = 12.2. What is the capacitance?

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Asked Jul 30, 2019
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The figure shows a parallel-plate capacitor with a plate area A = 2.52 cm2 and plate separation d = 7.95 mm. The bottom half of the gap is filled with material of dielectric constant κ1 = 13.7; the top half is filled with material of dielectric constant κ2 = 12.2. What is the capacitance?

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Expert Answer

Step 1

When capacitors are connected in series, the total capacitance is less than any one of the series capacitors’ individual capacitance. In this case, the upper and the lower capacitors are arranged in series so that equivalent capacitance is given by equation (4).

d/2
k2
d/2
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d/2 k2 d/2

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Step 2
C capacitence of a parallel plate
E permittivity of free space
A area of each plate
=
ksA
d
d seperation between plates
k dielectric constant
capacitence of upper half is
C
(2)
2k&A
d
capacitence of lower half is
2k360A
C2
(3)
d
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C capacitence of a parallel plate E permittivity of free space A area of each plate = ksA d d seperation between plates k dielectric constant capacitence of upper half is C (2) 2k&A d capacitence of lower half is 2k360A C2 (3) d

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Step 3
1
1
+
1
[Gand C in series]
С с,
C12
CC2
(4)
equ
GC2
2k,sgA 2k360A
C.
2k,SA 2k,6 use equation (2) and (3) in (4)
equ
d
d
2KK2A60
(5)
( p
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1 1 + 1 [Gand C in series] С с, C12 CC2 (4) equ GC2 2k,sgA 2k360A C. 2k,SA 2k,6 use equation (2) and (3) in (4) equ d d 2KK2A60 (5) ( p

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