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Parallel Plate Capacitor
• device to store charge
– (also stores energy)
• connect capacitor to battery (V)
– plates become oppositely
charged +Q/-Q
+Q -Q
A
charge  potential difference
Q = CV
d
+
• constant of
proportionality is called
the capacitance C
-
12 V
The 750mF capacitor in a camera flash attachment is
charged up by a 6V battery. What Q does it store?
Q=CV = (6 volts)(75010-6 coul/volt)
= 4.510-3 coul
ConcepTest
A one millifarad capacitor has is has a
voltage of 100 V across it. What is the
total charge on the capacitor?
1) Zero
2) 0.1 C
3) 10 C
4) 100,000 C
ConcepTest
A one millifarad capacitor has is has a
voltage of 100 V across it. What is the
total charge on the capacitor?
1) Zero
2) 0.1 C
3) 10 C
4) 100,000 C
What is the charge on the positive plate
of the capacitor?
1) Zero
2) 0.05 C
3) 0.1 C
4) 0.2 C
Capacitors
Capacitance depends only on geometry
Example: parallel-plate capacitor
A = area of plate
d = separation between plates
C  e0 A
d
k 
e0 = 8.85 x 10-12 C2/N-m2
(permittivity of free space)
1
4 e0
How big is 1 Farad?
C  e0 A
d
• choose d = 1 mm
–find the area A for such a
capacitor
A = C (d/e0)
= (1 F) (0.001 m) / (8.85x10-12)
= 1.1 x 108 m2 = 110 km2
This capacitor is as big as a city!!!
• Capacitor C1 is connected across a
battery of 5 V. An identical capacitor
C2 is connected across a battery of 10
V. Which one has the most charge?
1) C1
2) C2
3) both have the same charge
4) it depends on other factors
What must be done to a capacitor in
order to increase the amount of
charge it can hold (for a constant
voltage)?
1) increase the area of the plates
2) increase the separation between the plates
3) decrease the separation between the plates
4) decrease the area of the plates
5) either (1) or (2)
6) either (1) or (3)
7) either (2) or (4)
8) either (3) or (4)
Q = CV
C  e0 A
d
Increasing the separation, d, of the
two charged plates of a capacitor,
after it has been disconnected from
its battery, will
(1) Increase the charge on the plates.
(2) Increase the capacitance of the plates.
(3) Increase the voltage across the plates.
(4) None of the above.
• A parallel-plate capacitor initially has
a potential difference of 400 V and is
then disconnected from the charging
battery. If the plate spacing is now
doubled (without changing Q), what is
the new value of the voltage?
1) 400 V
2) 200 V
4) 800 V
5) 1600 V
3) 100 V
Inserting a dielectric material between
two charged parallel conducting
plates, originally separated by air
and now disconnected from the
battery, will
(1) Increase the charge on the plates.
(2) Increase the capacitance of the plates.
(3) Increase the voltage across the plates.
(4) None of the above.
Answers to Multiple Choice Questions
2) 0.1 C
3) 0.1 C
2) C2
6)
(3)
4)
(2)
either (1) or (3)
Increase the voltage across the plates.
800 V
Increase the capacitance of the plates.

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