PHY 184
Spring 2007
Lecture 12
Title: Capacitor calculations
1/29/07
184 Lecture 12
1
Announcements
 Homework Set 3 is due tomorrow morning at 8:00 am.
 Midterm 1 will take place in class next week on Thursday,
February 8.
 Practice exam will be posted in a few days.
 Second half of this Thursday’s lecture: review.
1/29/07
184 Lecture 12
2
Review of Capacitance
 The definition of capacitance is
q
C
V
1C
1F
1V
 The unit of capacitance is the farad (F)
 The capacitance of a parallel plate capacitor is given by
C
 Variables:
A is the area of each plate
d is the distance between the plates
1/29/07
184 Lecture 12
0 A
d
3
Cylindrical Capacitor
 Consider a capacitor constructed of two collinear conducting
cylinders of length L.
 The inner cylinder has radius r1 and
the outer cylinder has radius r2.
 Both cylinders have charge per
unit length  with the inner cylinder
having positive charge and the outer
cylinder having negative charge.
 We will assume an ideal cylindrical capacitor
• The electric field points radially from the inner cylinder to the outer
cylinder.
• The electric field is zero outside the collinear cylinders.
1/29/07
184 Lecture 12
4
Cylindrical Capacitor (2)
 We apply Gauss’ Law to get the electric field between the two cylinder
using a Gaussian surface with radius r and length L as illustrated by the
red lines
 
 0  E  dA  q
 0EA  L where A  2 rL
 … which we can rewrite to get an
expression for the electric field
between the two cylinders
1/29/07
184 Lecture 12

E
2 0 r
5
Cylindrical Capacitor (3)
 As we did for the parallel plate capacitor, we define the voltage
difference across the two cylinders to be V=V1 – V2.
r2
V1  V2   r
1
 
r
E  ds   r
2
1

dr
2 0r
 r2 


ln  
2 0  r1 
 The capacitance of a cylindrical capacitor is
q
C 
V
1/29/07
L
2 0 L


ln r2 / r1  ln r2 / r1 
2 0
Note that C depends on geometrical
factors.
184 Lecture 12
6
Spherical Capacitor
 Consider a spherical capacitor formed by two concentric conducting
spheres with radii r1 and r2
1/29/07
184 Lecture 12
7
Spherical Capacitor (2)
 Let’s assume that the inner sphere has charge +q and the outer sphere
has charge –q.
 The electric field is perpendicular to the surface of both spheres and
points radially outward
1/29/07
184 Lecture 12
8
Spherical Capacitor (3)
 To calculate the electric field, we use a Gaussian surface
consisting of a concentric sphere of radius r such that r1 < r < r2
 The electric field is always perpendicular to the Gaussian surface so
 … which reduces to
E
1/29/07
q
4 0 r
…makes sense!
2
184 Lecture 12
9
Spherical Capacitor (4)
 To get the electric potential we follow a method similar to the one we
used for the cylindrical capacitor and integrate from the negatively
charged sphere to the positively charged sphere
r1
r1
r2
r2
V   Edr  
q  1 1
dr 
 
2

4 0 r
4 0  r1 r2 
q
 Using the definition of capacitance we find
q
q
4 0
C 

V
q  1 1  1 1
    

4  r r   r r 
0
1
2
1
2
 The capacitance of a spherical capacitor is then
r1r2
C  4 0
r2  r1
1/29/07
184 Lecture 12
10
Capacitance of an Isolated Sphere
 We obtain the capacitance of a single conducting sphere by
taking our result for a spherical capacitor and moving the
outer spherical conductor infinitely far away.
 Using our result for a spherical capacitor…
C
q

V
q
4 0

q  1 1  1 1
    

4 0  r1 r2   r1 r2 
 …with r2 =  and r1 = R we find
C  4 0 R
1/29/07
…meaning V = q/40R
(we already knew that!)
184 Lecture 12
11
Example
 The “plates” of a spherical capacitor have radii 38 mm and
40 mm.
b=40 mm
a=38 mm
a) Calculate the capacitance.
b) Calculate the area A of a parallel-plate capacitor with the
same plate separation and capacitance.
d
A?
Answers: (a) 84.5 pF; (b) 191 cm2
1/29/07
184 Lecture 12
12
Clicker Question
 Two metal objects have charges of 70pC and -70pC,
resulting in a potential difference (voltage) of 20 V
between them. What is the capacitance of the
system?
A) 140 pF
B) 3.5 pF
C) 7 pF
D) 0
1/29/07
q 70 pC
C 
 3.5 pF
V 20 V
184 Lecture 12
13
Clicker Question
 Two metal objects have charges of 70pC and -70pC,
resulting in a potential difference (voltage) of 20 V
between them. How does the capacitance C change if
we double the charge on each object?
A) C doubles
B) C is cut in half
C) C does not change
The capacitance is the constant of proportionality between change and voltage. It depends
on the geometry not on the charge or voltage.
1/29/07
184 Lecture 12
14
Capacitors in Circuits
 A circuit is a set of electrical devices connected
with conducting wires.
 Capacitors can be wired together in circuits in
parallel or series
• Capacitors in circuits connected by wires such that the
positively charged plates are connected together and the
negatively charged plates are connected together, are
connected in parallel.
• Capacitors wired together such that the positively
charged plate of one capacitor is connected to the
negatively charged plate of the next capacitor are
connected in series.
1/29/07
184 Lecture 12
15
Capacitors in Parallel
 Consider an electrical circuit with three capacitors wired in
parallel
 Each of three capacitors has one plate connected to the
positive terminal of a battery with voltage V and one plate
connected to the negative terminal.
 The potential difference V across each capacitor is the
same.
.. key point for capacitors in parallel
 We can write the charge on each capacitor as …
q1  C1V q2  C2V q3  C3V
1/29/07
184 Lecture 12
16
Capacitors in Parallel (2)
 We can consider the three capacitors as one equivalent
capacitor Ceq that holds a total charge q given by
q  q1  q2  q3  C1V  C2V  C3V  C1  C2  C3 V
 We can now define Ceq by
Ceq  C1  C2  C3
q  CeqV
n
 A general result for n capacitors in parallel is
Ceq   Ci
i 1
 If we can identify capacitors in a circuit that are wired in
parallel, we can replace them with an equivalent capacitance
1/29/07
184 Lecture 12
17
Capacitors in Series
 Consider a circuit with three capacitors wired in series
 The positively charged plate of C1 is connected to the
positive terminal of the battery.
 The negatively charge plate of C1 is connected to the
positively charged plate of C2.
 The negatively charged plate of C2 is connected to the
positively charge plate of C3.
 The negatively charge plate of C3 is connected to the
negative terminal of the battery.
 The battery produces an equal charge q on each capacitor because the
battery induces a positive charge on the positive place of C1, which
induces a negative charge on the opposite plate of C1, which induces a
positive charge on C2, etc.
.. key point for capacitors in series
1/29/07
184 Lecture 12
18
Capacitors in Series (2)
 Knowing that the charge is the same on all three capacitors
we can write
 1
q
q
q
1
1
V  V1  V2  V3 


 q 
 
C1 C2 C3
 C1 C2 C3 
 We can express an equivalent capacitance Ceq as
q
V
Ceq
1
1
1
1



Ceq C1 C2 C3
 We can generalize to n capacitors in series
n
1
1

Ceq i 1 Ci
 If we can identify capacitors in a circuit that are wired in series, we can
replace them with an equivalent capacitance.
1/29/07
184 Lecture 12
19
Clicker Question
 C1=C2=C3=30 pF are placed in series. A battery supplies 9 V. What is the
charge q on each capacitor?
A) q=90 pC
B) q=1 pC
C) q=3 pC
D) q=180 pC
Ceq = 10 pF
Answer: 90 pC
1/29/07
184 Lecture 12
20
Clicker Question
 C1=C2=C3=1 pF are placed in parallel. What is the voltage of the battery
if the total charge of the capacitor arrangement, q1+q2+q3, is 90 pC?
A) 180 V
B) 10 V
C) 9 V
D) 30 V
Ceq = 3 pF
Answer: 30 volts
1/29/07
184 Lecture 12
21