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PHYS2012
EMP Workshop 1
A parallel plate capacitor has a square plate area of 0.200 m2 and a plate separation of
0.0100 m. The potential difference between the plates in air is 3.00 kV.
o = 8.8510-12 F.m-1
(1)
Sketch the capacitor and label it with the values of A, d and V0.
(2)
What do the following symbols represent?
Show that their values are correct. Explain why.
E0 = 3.00105 V.m-1
C0 = 1.7710-10 F
Q0f = 5.3110-7 C
0f = 2.6610-6 C.m-2
D0 = 2.6610-6 C.m-2
P0 = 0
A dielectric sheet of thickness 5.0010-3 m is inserted midway between the plates of
the charged capacitor that is not connected to a battery. After the insertion of the
dielectric, the potential difference between the plates is 2.00103 V.
(3)
Sketch the capacitor showing the distribution of the free and bound charges for
the plates and dielectric. On another sketch, show the field lines for E , D , P .
(4)
What do the following symbols represent?
Show that the following values are correct. Explain why.
t = 5.0010-3 m
V1 = 2.00103 V
1f = 2.6610-6 C.m-2
r = 3 need to integrate along a straight line from between the plates
t
r 
V    E dl
V
d  t  1
E0
e = 2
C1 = 5.3110-10 F
Q1f = 5.3110-7 C
E1 = 1.00103 V.m-1
D1 = 2.6610-6 C.m-2
1b / 1f = 2/3
1b = 1.77105 C.m-2
5
P1 = 1.7710 C.m-2
p2/em/wks1.doc
1
A parallel square plate capacitor has a plate area of 0.200 m2 and a plate separation of
0.0100 m is completely filled with a dielectric, r = 3.00 and is connected to a 2.00 kV
battery. From part (4)
C1 = 5.3110-10 F
V1 = 2.00103 V
(5)
What do the following symbols represent?
Show that the following values are correct. Explain why.
Q1f = 1.0610-6 F
U1 = 1.0610-3 J
The dielectric is partially removed from the dielectric so that it now only fills half the
space between the plates while the battery is still connected to the capacitor.
(6)
Show that the following values are correct. Explain why.
 A
C2  0 ( r  1)
2d
C2 = 3.5410-10 F
Q2f = 7.0810-7 C
U2 = 7.0810-4 J
change in charge stored by capacitor
q = -3.5410-7 C
Energy changes of the battery
Ubattery = 7.0810-4 J
Changes in energy stored by capacitor
Ucap = -3.5410-4 J
Work done by external force in removing dielectric
W = + 3.5410-4 J
Explain the signs of the energy changes and work.
We have the same parallel plate capacitor filled with the dielectric as in part (5). With
the battery still connected between the plates, the block of dielectric is withdrawn in a
direction parallel to the plates until only a length x remains between the plates.
(7)
Show that the external force in removing the dielectric is independent of the
position, x, of the dielectric.
Ignore any edge effects.
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Solution
(1) add diagrams
area of plates, A = 0.200 m2
separation distance of plates, d = 0.0100 m
Vo = 3.00103 V
o = 8.8510-12 F.m-1
(2) add diagrams
electric field between plates of capacitor
Eo = Vo / d = (3103 / 10-2) V.m-1 = 3.00105 V.m-1
capacitance of air filled capacitor
Co = o A / d = (8.8510-12)(0.2) / 0.01 F = 1.7710-10 F
free charge stored on capacitor plates
Q0f = Co Vo = (1.7710-10)( 3103) C = 5.3110-7 C
charge density of free charges
0f = Q0f / A = (5.3110-7) / 0.2 C.m-2 = 2.6610-6 C.m-2
electric displacement
D0 = 0f = 2.6610-6 C.m-2
polarization
P0 = 0b = 0
p2/em/wks1.doc
D0 = 0 E0
no dielectric
3
thickness of dielectric
t = 5.0010-3 m
voltage across capacitor V1 = 2.00103 V
(3)
The dielectric is polarized when inserted between the plates and the bound (induced polarized) charges Qb effectively neutralize some of the surface charges Qf. This then
reduces the electric field strength in the dielectric and thus lowering the potential
difference between the plates since E  V. The free charges Qf on the conductive
plates of the capacitor remains unchanged.
Gaussian surface 1
+Qf
top
+ + + + + + + + + + + + + + + + +
(d-t)/2
-Qb
-
-
-
-
-
-
d
t
+Qb
+
+
+
+
+
+
+
(d-t)/2
- - - - - - - - - - - - - - - - - - - - -
-Qf
bottom
Gaussian surface 2
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
- +
- +
- +
- +
- +
-
E
p2/em/wks1.doc
 f
b b
+
+
- +
+
+
+
- +
+
+
+
- +
+
+
+
- +
+
+
+ - +
+
P
o
- 
f
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
- +
- +
- +
- +
- +
D
-
o
4
(4)
add diagrams
thickness of dielectric, t = 5.0010-3 m
voltage across plate,
V1 = 2.00103 V
Free charges – no changes
1f = 2.6610-6 C.m-2
Eo = 3.00105 V.m-1
The potential difference, V1 across the plates can be found by integrating along a
straight line path from the bottom plate to the top plate
V1  Vtop  Vbottom   
top
bottom
E  ds  E0 (d  t )  E1t

t 
V1  E0  d  t  
r
r 

t
V
t
 1 d t
r 
V
 r E0
1
d t
E0
dielectric constant
r = 3
E1 
E0
electric susceptibility
e = r - 1 = 2
capacitance of dielectric filled capacitor
C1 = r o A / d = (3)(8.8510-12)(0.2) / 0.01 F = 5.3110-10 F
free charge stored on capacitor plates – does not change
Qf = 5.3110-7 C
electric field inside dielectric
E
3.00  105
E1  0 
V.m-1  1.00  105 V.m-1
r
3
electric displacement does not change – only depends upon free charges
D1 = 1f = 2.6610-6 C.m-2
surface charge density for bound charges

1
 1
1b  1 f 1     2.66  106  1   C.m-2  1.77  106 C.m-2
 3
 r 
1b
1 2
 1 
1 f
r 3
polarization


P1  D1   0 E1  2.66  106  8.85  1012 1  105  C.m-2  1.77  106 C.m-2
P1   1b  1.77  106 C.m-2
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(5) add diagrams
area of plates, A = 0.200 m2
separation distance of plates, d = 0.0100 m
r = 3.00
o = 8.8510-12 F.m-1
C1 = 5.3110-10 F
V1 = 2.00103 V
charge stored on capacitor plates
Q1f = C1 V1 = 1.0610-6 F
energy stored by capacitor
1
U1  C1 V12  (0.5)(5.31  1010 )(2  103 ) 2 J  1.06  103 J
2
(6) add diagrams
V2 = V1 = 2.00103 V
battery still connected
Capacitors in series C = C1 + C2
C2 
0 A
2d
( r  1)
C2 = 3.5410-10 F
charge stored on capacitor plates
Q2f = C2 V2 = (3.5410-10 ) (2.00103) = 7.0810-7 F
Energy stored by capacitor
1
U 2  C2 V2 2  (0.5)(3.54  1010 )(2  103 ) 2 J  7.08  10 4 J
2
change in energy stored by capacitor
Ucap  U2  U1  3.54 104
J
change in charge stored by capacitor
q = Q2f – Q1f = -3.5410-7 C
The charge on the plates of the capacitor decreases  charge transferred to battery
Ubattery  qV2  (3.54 107 )(2 103 ) J  7.08 104 J
Work is done by the external force in withdrawing the dielectric
W  Ucap  Ubattery    3.54  104  7.08  104 J  3.54  104 J
p2/em/wks1.doc
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W > 0  the external force must overcome the attraction between the capacitor plates
and the dielectric in removing it.
C  V = constant  Ucap  Q   Ubattery 
(7) add diagrams
Square plate capacitor A = 0.200 m2
and A = L2
L  A  0.2 m  0.4472 m
x = L/2 = 0.2236 m
V3 = 2.00103 V
Can consider the capacitor as two capacitors in series (C = C1 + C2)
C3 
 0 L( L - x )
d

r 0 L x
d

0 L
d
 L  x  r x
The energy stored by the capacitor decreases (energy stored in the electric field) is
1
 LV 2
U 3cap  C3 V32   0 3  L  x   r x 
2
2d
We also have to consider the change in the energy stored by the battery – energy
stored by battery increases
(W = q V, q = C V)
U 3battery  qV3  C3 V3 V3  C3 V32
 0 LV32
 L  x  r x
d
The total change in energy of the battery and capacitor is
U 3battery 
U 3  U 3cap  U 3battery
U3 
 0 LV32
2d
L  x  r x
Differentiating with respect to x gives the force acting on the dielectric
F
dU 3  0 LV32

 r  1
dx
2d
which is independent of x.
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