الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Capacitors and capacitance
q q
C = =
(6.5)
V qd A
ε ο
ε ο
A
∴ C =
(6.6)
d
Notice that the capacitance of the parallel plates capacitor is depends on the
geometrical dimensions of the capacitor.
The capacitance is proportional to the area of the plates and inversely
proportional to distance between the plates.
المعادلة (6.6)
تمكننا من حساب سعة المكثف من خلال الأبعاد الهندسية له،‏ حيث أن سعة
المكثف تتناسب طرديا ً مع المساحة المشتركة بين اللوحين وعكسيا ً مع المسافة بين اللوحين.‏
Example 6.1
An air-filled capacitor consists of two plates, each with an area of
7.6cm 2 , separated by a distance of 1.8mm. If a 20V potential difference
is applied to these plates, calculate,
(a) the electric field between the plates,
(b) the surface charge density,
(c) the capacitance, and
(d) the charge on each plate.
Solution
V 20
4
(a) E = = = 1.11×
10 V m
− 3
d 1.8×
10
(b) σ
(c) C
= ο
ε E =
ε A
=
d
= ο
(d) q = CV = (3.74×
10
−12
4
−8
2
( 8.85×
10 )(1.11×
10 ) = 9.83×
10 C m
×
1.8×
10
×
−12
−4
( 8.85 10 )(7.6 10 )
−12
= 3.74×
10
−3
−12
)(20) = 7.48×
10
−11
C
F
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