Electric devices function through storing and manipulating the flow of charges.
 This is achieved through several types discrete circuit components.
 Capacitors can store electric charge (and therefore electric potential energy) for a certain period of time.

Two
conductors
with equal
but opposite
charges
Now turn the problem around …
Given a potential difference (a voltage)
between two conductors, what is the charge
on them?
Electric
field
Potential
difference
between
the
charges
Q
C
V
(C/V=F)
Assume a
charge q on
the plates
Calculate the
e-field using
Gauss’ Law
 
 0  E  dA  q
 0 EA  q
Calculate
the
potential
 
V f  Vi    E  ds
Find the
capacitance
f
i

V   Eds

C
q
V
Increased area results in
increased charge storage for
a constant voltage.
Capacitance is proportional to area
+Q
+
V
E
-
V
-Q
V  Ed  V
+
+Q
E
-Q
-
V   Ed   V
The potential difference between the battery terminals and the
plates will create a field and charges will flow from the battery
to the plates until the potential difference is zeroed.
Capacitance is inversely proportional to the
distance between the plates.
Surface charge density on the plates:
The field (assuming
d<<A1/2):
The potential difference:
The capacitance:
Q

A

Q

E
0 0 A
Qd
V  Ed 
0 A
Q 0 A
C

d
V
E
2k e 
r
Q  l
b
b
a
a
Vb  Va    Edr  2ke 
dr
b
 2ke ln 
r
a
C
Q

V
C
l
l
b
2ke ln  
a
b
2ke ln 
a
1
C

l
b
2ke ln 
a


A battery drives
charges to accumulate
on the capacitor plates.
A switch controls the
flow of charge.
C,Q
+
V
Q  Q1  Q2
V  V1  V2
Q1  C1V1  C1V
Q2  C2 V2  C2 V
Q  Ceq V
Ceq  C1  C2
For parallel capacitors
n
Ceq   C j
j 1
All are parallel connections
This is not
V  V1  V2
Q  Q1  Q2
V1 
Q1 Q

C1 C1
V2 
Q2 Q

C2 C2
V 
Q
Ceq
1
1
1


Ceq C1 C2
For series capacitors
n
1
1

Ceq j 1 C j
C
0 A
d a
Ceq = ?
1
1
Rank the charges.
C1 = 3C
C2 = C
C3 = 5C
Rank the voltages.
∆
∆
∆
2



Since capacitors can carry charge after a battery
is disconnected, they can store energy.
To calculate this energy, consider the work done to
move a small amount of charge from one plate of
the capacitor to the other.
As more and more charges are moved, the work
required increases.
q
dW  Vdq  dq
C
Q
Q
q
1
Q2
W   dq   qdq 
C
C0
2C
0
Q2 1
1
U
 QV  C (V ) 2
2C 2
2
Consider a simple parallel-plate capacitor whose plates
are given equal and opposite charges and are separated
by a distance d. Suppose the plates are pulled apart until
they are separated by a distance D > d. The electrostatic
energy stored in the capacitor is
1. greater than
2. the same as
3. smaller than
before the plates were pulled apart.


Defined as stored energy per unit volume.
Consider the parallel plate capacitor.
1
U  ( 0 Ad ) E 2
2
1
U  C (V ) 2
2
0 A
d
(Ed )
2
U
2 ( 0 Ad ) E
uE 

Vol
Ad
1
uE   0 E 2
2
1
2
Dielectrics are insulators like glass, rubber or wood.
 They are used to increase the charge storage capacity and/or maximum voltage in capacitors while providing mechanical support.
 They are characterized by a dielectric constant, , and are limited by their dielectric strength.

As we apply a voltage to the plates, the dielectric in between is polarized.
 The dielectric constant is a measure of the polarizability of the material.
 This creates a field in the opposite direction to the applied field and therefore reduces the effective voltage or under constant voltage allows a larger charge to be collected on the plates.  The dielectric strength is the maximum electric field the material can withstand before breaking down (becoming a conductor).

V 
V0

Q0
Q0
C

V
V0
C  C0
A parallel-plate capacitor is attached to a
battery that maintains a constant potential
difference V between the plates. While the
battery is still connected, a glass slab is
inserted so as to just fill the space between the
plates. The stored energy
1. increases.
2. decreases.
3. remains the same.
0 A


C

C0
1 3   1  1 3 d
1 3  2 3
Capacitors are geometries that can hold charges.
 Use Gauss’ Law symmetries to calculate capacitance.
 Series and parallel connections.
 Dielectrics increase capacitance.


Reading Assignment
 Chapter 27 – Current and Resistance

WebAssign: Assignment 4