Page 6 - PRE-U STPM PHYSICS TERM 2

Physics Term 2 STPM Chapter 13 Capacitors

3. The electric field E between the plates is also given by

V
E = ………
d
Q V
Equating and , =
ε A d
0
–12
–1
Capacitance, C = Q = ε A where ε = 8.85 × 10 F m is the permittivity of free space between
0
0
13 the plates V d 13
Example 1

A capacitor is formed using two parallel metal plates, each measuring 15 cm × 20 cm separated by
a distance of 0.50 cm.
(a) Find the capacitance of the capacitor.
(b) If the plate separation is increased to 1.0 cm, what is the new capacitance?
(c) The new capacitor is connected to a battery of 12 V. What is the charge in the capacitor?

Solution:

ε A
0
(a) Capacitance, C = —–
d
(8.85 × 10 )(0.15 × 0.20)
–12
= F
(0.50 × 10 )
–2
= 5.31 × 10 F
–11
(b) Capacitance, C ∝ 1 , when d’ = 2d,
d
1
C’ = C
2
= 2.66 × 10 F
–11
(c) Charge, Q’ = C’V
= (2.66 × 10 )(12)
–11
= 3.19 × 10 C
–10

Quick Check 2

1. A parallel-plate capacitor has a capacitance of (i) What is the maximum potential
6.0 nF. The plates are separated by a distance difference that can be applied across
of 2.0 mm. What is the area of each plate? the capacitor?
(ii) What is the maximum charge that
2. The plates of a parallel-plate capacitor are can be stored in the capacitor?
separated by a distance of 5.00 mm. The area

2
of each plate is 0.040 m . 3. The capacitor used in the flash-light system
(a) Find the capacitance of the capacitor. of a camera has a capacitance of 24 mF. If the
(b) Insulation of the air between the capacitor is charged by a constant current of
capacitor breaks down if the electric 1.5 mA from a 6.0 V battery, what is the time
field between the plate exceeds taken to charge the capacitor to a potential
–1
6
3.0 × 10 V m . difference of 6.0 V?
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