Transcript Capacitors II

Capacitors II
Today’s plan
 Revision and some problems
 Energy storage in the capacitors
 Capacitors with dielectric material
Summary of important capacitor geometries
• The definition of the capacitance relates Q to V via C:
C
Q
V
• The capacitance depends on the geometry:
-Q
A
++++
d
-----
r
+Q
-Q
C
o A
d
In SI unit system:
a
a
b
b
L
Cylindrical
Parallel Plates
+Q
C
2 o L
b
ln  
a
Spherical
ab
C  4 o
ba
C has units of “Farads” or F (1F = 1C/V)
 o has units of F/m
• In a capacitor Charges will occupy
the faces facing each other.
• Fringing field is prominent if
L~d
• Parallel combination
Ceq  n Cn
• Series combination
1
1
 n
Ceq
Cn
Problem 8
Problem 8
 0 a x
C 
 x 
d 1 

d 

C 
 0 a x
d  x sin  
 0 a x
C 
d  x 
a
C
0
C
 0a 
1
x 
1 
 dx
d 
d 
 0a 
a 
1 

d  2d 
2
Problem 10
Rule
• Identify the points between which Ceq has to
be calculated.
• Connect the battery and send +Q from +ive
and –Q from –ive terminal
• Write charges appearing on each plate and
potential on each capacitor.
• If the plates form an isolated system total
charge is zero.
No charge
Isolated
system
Total charge
zero
Problem 10
No charge
Problem 10
C1 = C3 = C4 = C5 = C
C
C
y
x
C
Ceq = C
C
Electrostatic energy
stored in a capacitor
Work done by the external agent to charge the
capacitor is equal to electrostatic energy stored in
the capacitor.
Capacitor is connected
to a battery
Battery is doing work against the
electric field of plates
Battery is doing work against the
electric field of plates
Battery is doing work against the
electric field of plates
Battery is doing work against the
electric field of plates
•At some instance, let the charge on the
capacitor is q  V = q/C
+
+
+
-
Energy storage
• Work done in transferring next piece of
charge dq  dW = V dq
Q
2
q
1Q
1
2
W   dq 
 CV
c
2
C
2
0
Equal to electrostatic energy stored in the
capacitor
Where does this energy
reside??
• In the volume between the plates
• More specifically, in the electric field
that is present between the plates.
Energy density
1
2
CV
1
2
uA 
 0E 2
Ad
2
Empty space can contain energy
Capacitors with Dielectric
Empirical observation
• Capacitance increases
• It is good, since hard to make big
capacitors.
The role of dielectric
  
E  E0  Ein
E  E0  Ein
E0
E
ke
Case 1: Capacitor is not disconnected
from battery
+
+
+
-
Now the
dielectric
slab is
inserted
Capacitor is not disconnected from
battery
q
E
0 A
+
+
+
-
q
E 
 0 Ke A
But
E  E
q  ke q
Capacitor is not disconnected from
battery
q  ke q
+
+
+
+
+
+
-
Charge
will
increase
Capacity
increases
To be noted….
• Battery maintains a constant V
between the plates.
• Electric field inside the capacitors must
be the same.
• Battery moves additional charge q =
q´-q
• Capacity increases
Case 2: An empty capacitor is charged
and then disconnected from the battery.
Case 2: An empty capacitor is charged
and then disconnected from the battery.
Now filled with
dielectric
material
Case 1: An empty capacitor is charged
and then disconnected from the battery.
• Dielectric reduces the
strength of the electric
field.
•E = E0/k (always k >1)
V decreases and C increases.
Capacities with the dielectric
• Parallel plate capacitor
• Spherical capacitor
• Cylindrical capacitor
C
 0 ke A
d
ab
C  4 0 ke
ba
L
C  2 0 ke
b
ln 
a
Problem 20
• Find the capacitance of the capacitor
C´
 0 k 2 A / 2  0 k3 A / 2
C 
d
d
0 A / 2
k2  k3 
d
Ceq  C  C
 0 A  k1
k 2 k3 
 

Ceq 
2d  2 k 2  k 3 
P-14-15 P698
• Show that force per unit area acting on
either
capacitor
plate
(Electrostatic
pressure) is given by
1
2
F  0E
2
1 Q2 1 Q2 x
U

2 C 2 0 A
Force by which the plates will
attract each other
dU
Q2
F

dx 2 0 A