الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Capacitors and capacitance
Example 6.11
A Parallel plate capacitor of area 0.64cm 2 . When the plates are in
vacuum, the capacitance of the capacitor is 4.9pF.
(a) Calculate the value of the capacitance if the space between the
plates is filled with nylon (κ=3.4).
(b) What is the maximum potential difference that can be applied
to the plates without causing discharge (E max =14×10 6 V/m)?
(a)
Solution
C = κC = 3.4×4.9 = 16.7pF
o
(b) V max =E max ×d
To evaluate d we use the equation
d
ε A
=
C
= ο
o
× ×
4.9×
10
×
−12
−5
8.85
10 6.4 10
−4
= 1.16×
10
−12
V max = 1×10 6 ×1.16×10 -4 =1.62×10 3 V
m
Example 6.12
A parallel-plate capacitor has a
capacitance C o in the absence of
dielectric. A slab of dielectric material of
dielectric constant κ and thickness d/3 is
inserted between the plates as shown in
Figure 6.16. What is the new capacitance
when the dielectric is present?
1/3d
K
2/3
d
d
Figure 6.16
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