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Concept Question: Dielectric
A parallel plate capacitor is charged to a total charge Q and the
battery removed. A slab of material with dielectric constant  is
inserted between the plates. The charge stored in the capacitor
+ + + + + + + + + + +

          
1. increases
2. decreases
3. remains the same.
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1
Concept Question: Dielectric
A parallel plate capacitor is charged to a total charge Q and the
battery removed. A slab of material with dielectric constant  is
inserted between the plates. The energy stored in the capacitor
+ + + + + + + + + + +

          
1. increases
2. decreases
3. remains the same.
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Concept Question: Dielectric
A parallel plate capacitor is charged to a total charge Q and the
battery removed. A slab of material with dielectric constant  is
inserted between the plates. The force on the dielectric
+ + + + + + + + + + +

          
1. pulls in the dielectric
2. pushes out the dielectric
3. is zero.
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2
Capacitors: Store Electric Charge
Capacitor: Two isolated conductors
Equal and opposite charges Q
Potential difference V between them.
C
Q
V
Units: Coulombs/Volt or Farads
C is always positive.
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Finding Capacitance
1. Assume Q and determine V in terms of Q
2. Assume V and determine Q in terms of V
We will discuss the former method:
1. Choose a suitable coordinate system.
2. Assume +Q and –Q charges in the conductors.
3. Determine E and V.
4. Find C.
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3
Parallel-Plate Capacitor
Dielectric

1
2
Plate area
S

E
+ + + + + + + + +
        
x
d
0
Plates 1 and 2 carry charges +Q and –Q uniformly distributed on them.
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Parallel-Plate Capacitor
Consider homogeneous dielectric
between the plates.
Neglect fringing at the edges.
Dielectric

1
2
x
Plate area
S
d

E
+ + + + + + + + +
        
0
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Parallel-Plate Capacitor
Determine E
S 
Dielectric
 
Q
Q
 E  S  aˆ x   
aˆ
S

S x

Determine V

1
d
Q 
Qd
V   E  d l   
aˆ x   dx aˆ x 
2
0
S
 S 
Determine C
C
1
2
x
Plate area
S
d

E
+ + + + + + + + +
        
0
Q S

V
d
Dielectric constant r can be measured by measuring C!
r 
C
C0
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Energy Stored in Capacitor
Start Charging Capacitor
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Energy to Charge Capacitor
+Q
Q
1. Capacitor starts uncharged.
2. Carry +dq from bottom to top. Now top has charge q = +dq,
bottom –dq.
3. Repeat.
4. Finish when top has charge q = +Q, bottom –Q.
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Work Done Charging Capacitor
Potential difference is: V  q / C
So work done to move dq is:
dW  dqV  dq
q 1
 q dq
C C
Total energy to charge to q = Q:
1 Q
q dq
C 0
Q2
W
2C
W   dW 
+Q
Q
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Work Done Charging Capacitor
Since, C 
W
Q
V
Q2 1
1
 QV  CV 2
2C 2
2
Where is the energy stored?
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Energy Stored in Capacitor
Energy stored in the E field!
1
 E 2 dv

2
1
Q2
1  Q 2 Sd
   2 2 dv 
2  S
2  2S 2
Q2  d  Q2 1

 QV


2   S  2C 2
WE 
Dielectric

1
2
x
Plate area
S

E
+ + + + + + + + +
        
d
0
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Coaxial Capacitor
Q
+Q
 Conductors 1 and 2 carry charges +Q and –Q uniformly
distributed on them.
 Consider homogeneous dielectric between the plates.
 Neglect fringing at the edges.
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Coaxial Capacitor
Determine E
 
Q    E  dS   E 2 L

E
Q
2 L
Q
+Q
aˆ 
Determine V

1
a

Q
V   E  d l   
aˆ    d  aˆ 
2
b 2 L


Q
b

ln
2 L a
Determine C
C
Q 2 L

b
V
ln
a
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Spherical Capacitor
Q
+Q
Inner and outer spheres carry charges +Q and –Q uniformly
distributed on them.
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Spherical Capacitor
Determine E
 
Q    E  dS   Er 4 r 2

E
Q
Q
aˆ r
4 r 2
+Q
Determine V

1
a
Q

V   E  d l   
aˆ  dr aˆ r
2
b 4 r 2 r 


Q 1 1


4  a b 
Determine C
C
Q
4

1
1
V 

 a b 
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Concept Question: Changing Dimensions
A parallel plate capacitor has plates with equal and opposite charges
Q, separated by a distance d, and is not connected to a battery. The
plates are pulled apart to a distance D>d. What happens?
1. V increases, Q increases
x
2. V decreases, Q increases
3.
4.
5.
6.
7.
8.
9.
V is the same, Q increases
V increases, Q is the same
V decreases, Q is the same
V is the same, Q is the same
V increases, Q decreases
V decreases, Q decreases
V is the same, Q decreases
+Q
1
2
d

E
+ + + + + + + + +
        
0
Q
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Concept Question: Changing Dimensions
A parallel plate capacitor has plates with equal and opposite charges
Q, separated by a distance d, and is connected to a battery. The
plates are pulled apart to a distance D>d. What happens?
1. V increases, Q increases
x
2. V decreases, Q increases
3. V is the same, Q increases
+Q
1
d
4. V increases, Q is the same
+ + + + + + + + +

5. V decreases, Q is the same
E
6. V is the same, Q is the same
        
0
2
7. V increases, Q decreases
Q
8. V decreases, Q decreases
9. V is the same, Q decreases
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Concept Question: Changing Dimensions
A parallel plate capacitor has plates with equal and opposite charges
Q, separated by a distance d, and is not connected to a battery. The
plates are pulled apart to a distance D>d. How does the final
electrostatic energy stored in the capacitor compare to the initial
energy?
x
1. The final stored energy is smaller
+Q
2. The final stored energy is large
1
d
+ + + + + + + + +

3. Stored energy does not change
E
2
        
0
Q
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