Transcript Ch 17 Part 2

Devices that can store electric charge
are called capacitors.
 Capacitors consist of 2 conducting
plates separated by a small distance
containing an insulator.
 Capacitors are used in electronics,
computers, camera flashes, and as
protectors of circuits for surges or
memory (RAM) of binary code.

Capacitors become quickly charged when a
voltage is applied to it in a circuit.
 One plate will be negatively charged and the
other plate will have an equal positive
charge.
 The charge, Q, depends on the potential
difference, V, applied to it. Q = CV. C is a
proportionality constant called capacitance
of the capacitor and is measured in
coulombs/volt or farad (F).
 Common capacitance ranges are from 1pF
(picoFarad) to 1µF (microFarad) (10-12 to 10-6).


The Capacitance, C, depends on the area
of the capacitor plates and the distance of
insulated (or air) separation.
A
C 0
d

Recall the permittivity of free space constant
has a value of 8.85 x 10-12 C2 /N*m2

(a) Calculate the capacitance of a
capacitor whose plates are 20cm x 3.0
cm and are separated by a 1.0mm air
gap. (b) What is the charge on each
plate if the capacitor is connected to a
12-V battery? (c) What is the electric
field between the plates?

(a) The area A = 20 cm x 3.0 cm = 60.0
cm2 = 6.0 x 10-3 m2 so the capacitance,
3 2

6
.
0
x
10
m 
12 2
2
  53 pF
C  (8.85x10 C / Nm )
3
 1.0 x10 m 

(b) The charge on each plate is Q=CV
Q  (53x10

12
F )(12V )  6.4 x10
10
(c) For a uniform electric field b/t the
12V
plates, E=V/d
4
E
3
1.0 x10 m
C
 1.2 x10 V / m
A charged capacitor stores energy equal
to the work done to charge it.
 Batteries charge capacitors by removing
charge from one plate and putting charge
on the other plate. This takes time.
 The more charge already on a plate, the
more work required to add more charge of
the same sign.
 Initially when a capacitor is uncharged, it
takes NO WORK to move the first bit of
charge over.

The work needed to move a small amount
of charge Δq, when a potential difference,
V, is across the plates, is ΔW=VΔq.
 Voltage increases during the charging
process from 0 to its final value Vf .
 Since the average voltage during this
process will be Vf /2 then the total work, W
to move all of the charge, Q at once is the
energy stored, U, in the capacitor:
U=energy=1/2 QV.
 Since Q=CV, then U=1/2QV=1/2CV2
=1/2Q2/C.


A camera flash unit stores energy in a
150µF capacitor at 200V. How much
electric energy can be stored?
1
(150 x10 6 F )( 200V ) 2  3.0 J
2
U = energy = ½ CV2 =
 Notice how the units work out: recall
1F=1C/V and 1V=1J/C.
 If this energy could be released in 1/1000s
the power output would be 3000W.

Though energy is not a substance in a
place, it is helpful to think of it as being
stored in the electric field between the
plates.
 Recall the electric field b/t 2 parallel
plates separated by a small distance is
uniform and related to electric potential,
V by V=Ed. Also recall capacitance, C

A
C 0
d

Therefore, energy stored, U in terms of the
electric field is found
1
1  0 A  2 2
2
U  CV  
( E d )
2
2 d 
1
U  0 E 2 Ad
2
The quantity Ad is the volume between the
plates where the electric field E exists.
 Dividing both sides by the volume, we get the
energy / volume or energy density Energy stored is

1
U  energydensity  0 E 2
2
proportional to
the square of
electric field.

Please do Ch 17 Rev p 524 #s 30, 32, 33,
35, 36, 37, 43, 44, & 45