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Electrostatics #5
Capacitance
Capacitance
I. Define capacitance and a capacitor:
Capacitance is defined as the ability of an object to store charge.
A capacitor is an electronic component of circuits.
parallel
The simplest capacitor is one made of two ________________
metal
sheets, or plates.
Q
Q
Q  the amount of charge stored on the
capacitor. The parallel plates always
carry equal but opposite charges.
Note: Where there is a charge separation, there is also
electric
an ____________
field. The field exists between the


plates and points from _____ towards ____ .
electric
Also, where there is an electric field, there is also an ___________
potential . The _____ plate is the higher electric potential and the
___________


_____ plate is the lower electric potential.
lower electric potential
higher electric potential
direction of electric field
For a constant electric field, the change of electric potential is given by:
V  Ex cos
For the capacitor, this can be simplified to
V  Ed
This is just the magnitude of the electric potential between the two
plates.
V and the
The capacitance is defined through the electric potential _____
Q . The equation for capacitance is:
charge held on the plates, _____
Q  C V
next slide for definitions…
Q  C V
|V| = the electric potential between the two plates. When charge is
placed on a capacitor, an initial electric potential must be provided
by a battery.
Q = the electric charge stored on the capacitor
C = the capacitance of the device. The greater the capacitance, the
more charge stored on the device for a given voltage.
farad
The units of capacitance are given the special name of ____________
and have a symbol of ‘F’. What is the farad equivalent to?
Q
C
V
Q
C
V
charge

electric potential
coulomb C


volt
V
C
C
C2
1 farad  1 F  1  1  1
J
V
J
C
Ex. 1: A parallel plate capacitor is made of circular metal sheets placed 0.100
mm apart and has a capacitance of 1.00 mF. If air is used as the insulator
between the two metal plates, what is the maximum amount of charge that may
be stored on this capacitor? Air ceases to be an insulator when the electric field
is larger than 3 10 6 N
C
Q  CV
and
V  Ed
Q  CEd




6 N 
3
Q  1.00 10 F  3 10
0
.
100

10
m

C

6
Q  3.00 104 C

II. The capacitance of a parallel plate capacitor can be calculated from its
dimensions: The area of the overlap of the two sheets or plates and the distance
between the plates.
A = the area that the two surfaces overlap (one covers
the other)
A
d = the distance between the plates (plate separation)
area and
The capacitance, C, is proportional to the ________
d
distance between the
inversely proportional to the _______________
plates.
A
C
d
or
C
o A
d
The constant o is called the electric permittivity of free space, and the value
of o is:
 o  8.854 1012
C2
12 F
 8.854 10
2
N m
m
2
N

m
The permittivity constant is related to the coulomb constant, k  8.988 10
C2
9
The coulomb constant is derived from the force between charges, and the
permittivity constant is derived through Gauss’ Law. The actual relation is:
k
1
4  o
Ex. 2: A 1.00 mF capacitor is constructed with its metal plates set 0.100 mm
apart. If the plates are circular in shape, what is the diameter of the plates?
C
r
Cd
 o

o A
d

 o r 2
d
1.00 10 F 0.100 10 m
6

12 F 
  8.854 10

m

r  1.896 m  diameter  3.79 m
3
Ex. 3 A 1.00 F capacitor is constructed with square metal plates set 1.00 mm
apart. What is the length of a side for the metal plates?
C
s
o A
d
Cd
o


os2
d
s = the length of one side
of the square area
1.00 F 1.00 103 m

12 F 
 8.854 10

m

s  1.06 104 m  10.6 km  6.6 miles
III. Energy stored in a capacitor: Since the parallel plate capacitor has two
plates that are oppositely charged, there is energy stored in the electric
interaction between the two plates. This energy is stored in the electric field
between the two plates. The energy is:
2
Q
U  12 Q V  12 C V 
2C
U = the electric energy
stored in a capacitor.
2
Ex. 4: A parallel plate capacitor is made with an air gap of 0.0100 mm and
circular plates with a diameter of 3.25 cm. a. What is the capacitance of this
capacitor?
2
C
o A
d

 o r
d

12 F  3.25 10 m 

  8.854 10

m 
2


C
0.0100 103 m
2


2
 7.35 10 10 F
b. What is the maximum charge that may be placed on this capacitor? Let E
have a value of 3 10 N
6
C
Q  CV
and
V  Ed
Q  CEd



N

Q  7.35 10 10 F  3 106  0.0100 10 3 m
C

Q  2.20 108 C

c. What is the energy stored in this capacitor?

2.20  10 C 
Q

U
10

2
7
.
35

10
F
2C
8
2
2
 3.3110 7 J
d. What is the energy density between the plates of the capacitor?
U
energy
u

Vol. volume
U
u
 r 2h
u = the energy density, or energy per unit volume.

3.3110 J 

 3.25 10 m 
 0.0100 10 m 
 
2
7
2
2

3

J
u  39.8 3
m
Note: Alternate form to energy density! You do not need to memorize this
derivation…
C V
U
u

volume
Ad
1
2
u
u  oE
1
2
2
1
2
o A
Ed
d
Ad
2
2
u  12  o E 2

12 F 
6 N 
  8.854 10
 3 10

m 
C

1
2
2
J
 39.8 3
m