Page 5 - PRE-U STPM PHYSICS TERM 2

Physics Term 2 STPM Chapter 13 Capacitors

6. Since a conductor is able to store charges, the concept of capacitance can also be applied to a
conductor. The capacitance of a conductor is the ratio of the charge Q on the conductor to the
electric potential V of the conductor.
Charge on conductor
Capacitance of conductor = ————————————––—
Electric potential of conductor
Q
C = —
13 V 13

7. The unit of capacitance is farad (F).
The farad (1 F) is the capacitance of a capacitor that has a charge of one coulomb (1 C)
on each plate when the potential difference between the plates is one volt (1 V).
Capacitors with capacitance of a few µF are used in simple radio receiver circuits and large capacitors
capacitance of a few MF are used in electrical appliances such as washing machines.

Quick Check 1

1. A capacitor stores 24 mC of charge when Sketch a graph to show how the charge Q
the potential difference between the plates is in the capacitor varies with the potential
12 V. What is the capacitance of the capacitor? difference V across the capacitor.

2. A 50 µF capacitor is charged by connecting it 3. The capacitor of variable capacitance is
to a 6.0 V battery. connected to a 12 V supply. Its capacitance
What is the charge in the capacitor when the is 50 µF. After the capacitor is charged,
potential difference across the capacitor is the voltage supply is disconnected and the
(a) 3.0 V? capacitance is changed to 100 µF. What is the
(b) 6.0 V? new potential difference across the capacitor?

13.2 Parallel-Plate Capacitorsapacitors
C
Parallel-Plate
13.2
2016/P2/Q5
Learning Outcomes
Students should be able to:
• describe the mechanism of charging a parallel-plate capacitor
Q ε A
• use the formula C = to derive C = 0 for the capacitance of a parallel-plate capacitor
V d

1. Figure 13.5 shows a parallel-plate capacitor which consists of two
parallel metal plates each of area A and separated by a distance d d
in free space or vacuum. The capacitor is charged to a potential
difference V.
Q Q
E
2. The charge on each plate is Q. Using Gauss’s law,
ε Φ = ΣQ
0
ε (EA) = ΣQ V
0
Q
E = ………
ε A Figure 13.5
0

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