PHYS 1444
Lecture #5
Tuesday June 19, 2012
Dr. Andrew Brandt
•
•
Tuesday, June 19, 2012
Short review
Chapter 24
•
Capacitors and Capacitance
PHYS 1444 Dr. Andrew Brandt
1
Coulomb’s Law – The Formula
Q11
Q
Q22
Q
F
2
r
Formula
A vector quantity. Newtons
Q1Q2
F k
2
r
• Direction of electric (Coulomb) force (Newtons) is always
along the line joining the two objects.
• Unit of charge is called Coulomb, C, in SI.
• Elementary charge, the smallest charge, is that of an
electron: -e where
19
e  1.602 10
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
C
2
Vector Problems
• Calculate magnitude of vectors (Ex.
force using Coulomb’s Law)
• Split vectors into x and y components
and add these separately, using
diagram to help determine sign
• Calculate magnitude of resultant
|F|=(Fx2+Fy2)
• Use = tan-1(Fy/Fx) to get angle
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
3
Gauss’ Law
• The precise relation between flux and the enclosed charge is
r r Qencl
given by Gauss’ Law
Ñ
 E  dA 
e0
• e0 is the permittivity of free space in the Coulomb’s law
• A few important points on Gauss’ Law
– Freedom to choose surface
– Distribution of charges inside surface does not matter only total
charge
– Charges outside the surface do not contribute to Qencl.
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
4
Example 22-3: Spherical conductor.
A thin spherical shell of radius r0 possesses
a total net charge Q that is uniformly
distributed on it. Determine the electric field
at points (a) outside the shell, and (b) within
the shell. (c) What if the conductor were a
solid sphere? *q3
r r Qencl
E  dA 
Ñ

e0
Figure 22-11. Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net
charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to
determine Example 22–3.
Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know
the field must be radial, so E = Q/(4πε0r2).
b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must
be zero.
c. All the excess charge on a conductor resides on its surface, so these answers hold for a
solid sphere as well.
Key to these questions is how much charge is enclosed
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
5
Example 22-4: Solid sphere of charge.
An electric charge Q is distributed
uniformly throughout a nonconducting
sphere of radius r0. Determine the
electric field (a) outside the sphere
(r > r0) and (b) inside the sphere (r < r0).
*q4
Solution: a. Outside the sphere, a gaussian surface encloses the
total charge Q. Therefore, E = Q/(4πε0r2).
b. Within the sphere, a spherical gaussian surface encloses a
fraction of the charge Qr3/r03 (the ratio of the volumes, as the
charge density is constant). Integrating and solving for the field
gives E = Qr/(4πε0r03).
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
6
Example 22-5: Nonuniformly charged solid
sphere.
Suppose the charge density of a solid sphere is
given by ρE = αr2, where α is a constant. (a)
Find α in terms of the total charge Q on the
sphere and its radius r0. (b) Find the electric
field as a function of r inside the sphere.
Solution: a. Consider the sphere to be made of a series of
spherical shells, each of radius r and thickness dr. The volume
of each is dV = 4πr2 dr. To find the total charge: Q = ∫ρE dV
= 4παr05/5, giving α = 5Q/4πr05.
b. The charge enclosed in a sphere of radius r will be Qr5/r05.
Gauss’s law then gives E = Qr3/4πε0r05.
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
7
Electric Potential Energy
• Concept of energy is very useful solving
mechanical problems
• Conservation of energy makes solving complex
problems easier.
• Defined for conservative forces (independent of
path)
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
8
Properties of the Electric Potential
• What are the differences between the electric potential and
the electric field?
1 Q
– Electric potential (U/q)
V
4e0 r
• Simply add the potential from each of the charges to obtain the total potential
from multiple charges, since potential is a scalar quantity
r
1 Q
– Electric field (F/q)
E
4e0 r2
• Need vector sums to obtain the total field from multiple charges
• Potential for a positive charge is large near a positive charge
and decreases to 0 at large distances.
• Potential for the negative charge is small (large magnitude but
negative) near the charge and increases with distance to 0
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
9
Capacitors (or Condensers)
• What is a capacitor?
– A device that can store electric charge without letting the charge flow
• What does it consist of?
– Usually consists of two oppositely charged conducting objects (plates or
sheets) placed near each other without touching
– Why can’t they touch each other?
• The charges will neutralize each other
• Can you give some examples?
– Camera flash, surge protectors, computer keyboard, binary circuits…
• How is a capacitor different than a battery?
– Battery provides potential difference by storing energy (usually chemical
energy) while the capacitor stores charge but very little energy.
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
10
Capacitors
• A simple capacitor consists of a pair of parallel plates
of area A separated by a distance d.
– A cylindrical capacitors are essentially parallel plates
wrapped around as a cylinder.
• Symbols for a capacitor and a battery:
– Capacitor -||– Battery (+) -|i- (-)
Tuesday, June 19, 2012
Circuit
Diagram
PHYS 1444 Dr. Andrew Brandt
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Capacitors
• What do you think will happen if a battery is connected
(voltage is applied) to a capacitor?
– The capacitor gets charged quickly, one plate positive and the other
negative with an equal amount. of charge
• Each battery terminal, the wires and the plates are
conductors. What does this mean?
– All conductors are at the same potential.
– the full battery voltage is applied across the capacitor plates.
• So for a given capacitor, the amount of charge stored in the
capacitor is proportional to the potential difference Vba
between the plates. How would you write this formula?
Q  CVba
C is a property of a capacitor so does not depend on Q or V.
– C is a proportionality constant, called capacitance of the device.
PHYS 1444 Dr. Andrew Brandt
12
Normally use mF or pF.
– What is the unit? C/V or Farad (F)
Tuesday, June 19, 2012
Determination of Capacitance
• C can be determined analytically for capacitors w/ simple
geometry and air in between.
• Let’s consider a parallel plate capacitor.
– Plates have area A each and separated by d.
• d is smaller than the length, so E is uniform.
– For parallel plates E=s/e0, where s is the surface charge density.
r r
• E and V are related Vba   E  dl

b
a
• Since we take the integral from the lower potential point a to
the higher potential point b along the field line, we obtain
b
b
b s
b Q
Q
Qd
Q b
V


Edl
cos180


Edl

dl

dl

b

a

V

V

dl



• ba b a a
a
a e 0 a e 0 A e 0 A a e 0 A
e0 A
• So from the formula: C  Q  Q  e 0 A C only depends on the area
– What do you notice?
Tuesday, June 19, 2012
Vba
Qd e 0 A
PHYS 1444 Dr. Andrew Brandt
d
(A) and the separation (d) of
the plates and the permittivity
of the medium between
13 them.
Example 24 – 1
Capacitor calculations: (a) Calculate the capacitance of a capacitor whose
plates are 20 cm x 3.0 cm and are separated by a 1.0 mm 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? (d) Estimate the area of the
plates needed to achieve a capacitance of 1F, given the same air gap.
(a) Using the formula for a parallel plate capacitor, we obtain
e0 A
C

d

 8.85  1012 C 2 N  m2

0.2  0.03m2
12
2

53

10
C
N  m  53 pF
3
1  10 m
(b) From Q=CV, the charge on each plate is


Q  CV  53  1012 C 2 N  m 12V   6.4  1010 C  640 pC
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
14
Example 24 – 1
(c) Using the formula for the electric field in two parallel plates
Q
s
6.4  1010 C
4
4


1.2

10
N
C

1.2

10
V m
E 
e 0 Ae 0 6.0  103 m2  8.85  1012 C 2 N  m2
Or, since V  Ed we can obtain
12V
V
4

1.2

10
V m
E 
3
d 1.0  10 m
(d) Solving the capacitance formula for A, we obtain
e0 A
C
Solve for A
d
1F  1  103 m
Cd
8 2
2

10
m

100
km

A
e0
9  1012 C 2 N  m 2


About 40% the area of Arlington (256km2).
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
15
Example 24 – 3
Spherical capacitor: A spherical capacitor consists of
two thin concentric spherical conducting shells, of radius
ra and rb, as in the figure. The inner shell carries a
uniformly distributed charge Q on its surface and the
outer shell an equal but opposite charge –Q. Determine
the capacitance of this configuration.
Using Gauss’ law, the electric field outside a E  Q
4e 0 r 2
uniformly charged conducting sphere is
So the potential difference between a and b is
Vba  


b
a

b
a
r r
E  dl 
E  dr  

b
a
Q 1b
dr
Q 1 1
dr




 
 
2
a
4e 0 r 2
4e 0 r
4e 0  r ra 4e 0  rb ra
Q
 4e 0 rb ra
Q


r

r
C
 Q
a
b
ra  rb
V 4e  r r 
0 
b a 
Q
Q

b
r

Q  ra  rb 




4
e
r
r
0 
b a 

Thus capacitance is
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
16
Capacitor Cont’d
• A single isolated conductor can be said to have a
capacitance, C.
• C can still be defined as the ratio of the charge to absolute
potential V on the conductor.
– So Q=CV.
• The potential of a single conducting sphere of radius rb can
be obtained as
Q
Q 1 1
V 
where ra  
  
4
e
r
4e 0  rb ra 
0 b
• So its capacitance is C  Q / V  4e 0 rb
• Although it has capacitance, a single conductor is not considered to be
a capacitor, as a second nearby charge is required to store charge
Tuesday, June 19, 2012
PHYS 1444 Dr. Andrew Brandt
17
Capacitors in Series or Parallel
• Capacitors are used in many electric circuits
• What is an electric circuit?
– A closed path of conductors, usually wires connecting capacitors and
other electrical devices, in which
• charges can flow
• there is a voltage source such as a battery
• Capacitors can be connected in various ways.
– In parallel
Tuesday, June 19, 2012
and
in Series
PHYS 1444 Dr. Andrew Brandt
or in combination
18
Capacitors in Parallel
• Parallel arrangement provides the same
voltage across all the capacitors.
– Left hand plates are at Va and right hand
plates are at Vb
– So each capacitor plate acquires charges
given by the formula
• Q1=C1V, Q2=C2V, and Q3=C3V
• The total charge Q that must leave battery is then
 Q=Q1+Q2+Q3=V(C1+C2+C3)
• Consider that the three capacitors behave like a single
“equivalent” one
For capacitors in parallel the capacitance is the
 Q=CeqV= V(C1+C2+C3)
sum of the individual capacitors
• Thus the equivalent capacitance in parallel is
Ceq  C1  C2  19
C3
Capacitors in Series
• Series arrangement is more “interesting”
 When battery is connected, +Q flows to the left plate of
C1 and –Q flows to the right plate of C3
 This induces opposite sign charges on the other plates.
 Since the capacitor in the middle is originally neutral,
charges get induced to neutralize the induced charges
 So the charge on each capacitor is the same value, Q. (Same charge)
• Consider that the three capacitors behave like an equivalent one
 Q=CeqV  V=Q/Ceq
• The total voltage V across the three capacitors in series must be equal to
the sum of the voltages across each capacitor.
 V=V1+V2+V3=(Q/C1+Q/C2+Q/C3)
• Putting all these together, we obtain:
• V=Q/Ceq=Q(1/C1+1/C2+1/C3)
1
1
1
1



• Thus the equivalent capacitance is Ceq C1 C2 C3
PHYS 1444 Dr. Andrew Brandt
The total capacitance is smaller than the smallest C!!!
20