Can current flow in electric shock?
Yes. Transient current can flow in insulating medium
in the form of time varying displacement current.
This was an important discovery made
by Maxwell who could predict that
electromagnetic waves can be propagated
in vacuum.
Note 4: Capacitance and Dielectrics
Two conductor plates of area A separated by distance d
constitute a capacitor. Capacitors are able
to store electric energy. (Batteries store
chemical energy that can be converted to
electric energy.)
The electric field due to a surface charge on conductor
surafce was found using Gauss' Law
Q
σ
E =
(σ surface charge density)
=
ε0 ε0 A
V
Since E = (potentail gradient), we find
d
ε A
Q ε A
C

Q
= 0 V= CV , C
= = 0 Capacitance  =Farad 
d
V
d
V

d (m)
Dimensions of ε 0 : Farad
= Farad/m
2
A (m )
ε 0 means capacitance per unit length in vacuum.
V (volts)
+
+
+Q +
E
- -Q
+
-
+
-
+
-
Various Types of Capacitors
A
is small quantity because=
ε 0 8.85 × 10−12 F/m
d
is small. Dielectric materials (insulators) have permittivities larger than
ε 0 and the capacitance can be increased by filling the capacitor with a
Capacitance=
C ε0
dielectric material. (Stud finder detects the chnge in the cpacitance.)
If the sepration distance can be made small, a large capacitance can be
realized. In electrolytic capacitor, thin layer of oxide insulator is formed
on one conductor surface. Electrolytic capacitors have relatively large
capacitance but operate at low voltage with fixed polarity.
Oil capacitors are used for storing large amount of electric energy for
discharging in short time. (The capacitor for the STOR-M tokamak in
the Plasma Physics Laboratory can store energy of 300 kJ at 7 kV.)
Example 1. Cylindrical capacitor of length l , inner and outer conductor radii a and b
The electric field was found earlier,
1 Q1
Er =
2πε 0 l r
The potential difference between the outer and inner conductors is
a
1 Q a1
1 Q b
ln  
V =
dr
− ∫ Er dr =
−
=
b
2πε 0 l ∫b r
2πε 0 l  a 
We have assumed that the outer conductor is grounded (V = 0 ) . The capacitance is
C
=
2πε 0 l
Q
(F)
=
V ln ( b / a )
If the capacitor is filled with
dielectric material with a permittivity ε ,
Q
2πε l
= =
C
(F)
V ln ( b / a )
Example 2: Coaxial cable has inner conductor radius of 1 mm and
outer conductor radius of 3 mm. It is filled with Teflon which has a
a permittivity ε =2ε 0 . Find the capacitance per unit length of
the cable.
Sol. C / l is
C 2π × 2ε 0 2π × 2 × 8.85 × 10−12
= =
F/m = 5.06 × 10−11 F/m
l ln ( b / a )
ln 3
Example 3: Self-capacitance of conducting sphere
If a conducting sphere of radius a is charged with a charge Q, the
potential of the shpere is raised to
V =
1 Q
4πε 0 a
as found in Note 3. This defines the self-capacitance of the conducting
sphere
Q
= = 4πε 0 a (F)
C
V
Spherical Capacitor
Example 4. Spherical capacitor consists of inner spherical electrode
of radius a and outer spherical shell of radius b. Find the capacitance.
Sol. Let the inner conductor carry a charge Q and the outer conductor
−Q. From Gauss law, the electric field an be found as
1 Q
in the region between the electrodes a < r < b
Er =
2
4πε 0 r
The potential difference between the electrodes is
a
Q
b
4πε 0
∆V =− ∫ Er dr =−
∫
a
b
1
Q
dr
=
−
4πε 0
r2
The capacitance is C = 4πε 0
ab
(F)
b−a
a
Q
 1
−
=


 r  b 4πε 0
1 1
 − 
a b
Parallel and Series Connection of Capacitors
Parallel connection
The voltage is common. The charges are
=
Q1 C=
C2V
1V , Q2
The total charge is
Q = Q1 + Q2 = ( C1 + C2 ) V
Equivalent capacitance is
Ceq= C1 + C2 (parallel connection)
Series connection
The charge is common but the voltages differ.
=
V1
Q
Q
, V2
=
C1
C2
 1
1 
V = V1 + V2 = Q  +

 C1 C2 
1
1
1
1
Q
+
or =
Ceq= =
1
1
V
Ceq C1 C2
+
C1 C2
Example 5. Find the effective capacitance.
Sol. Use parallel and series formulae repeatedly.
Energy Storage in Capacitor
A capacitor C having a charge q is at a voltage V =
q
. If the charge is increaesd
C
by dq, the change in the energy is
1
= Vdq
=
dU
qdq
C
The energy stored in the capactitor is given by
1
=
U
C
1 Q2 1
1
2
=
=
qdq
=
CV
QV (J)
∫0
2 C
2
2
Q
V
dU =Vdq
dq
area
= VQ/2 (J)
Q
Energy Density
Example 6. Energy storage in a parallel plate capacitor with plate area A, separation
distance d . If it is charged to charge Q, the stored energy is
1 Q2 1 Q2
1 σ2
1
ε 0 E 2 × Volume (Ad )
=
=
=
U =
Ad
2 C
2 ε0 A / d 2 ε0
2
Therefore, an energy density
1
ε 0 E 2 (J/m3 )
2
is associated with electric field E (V/m).
u=
Example 6 : In the arrangement, find the energy stored in each capacitor.
Sol. The voltage in each capacitor is:
6
4
3 µ F and 2 µ F: V =
× 90 V =
× 90 V = 60 V
3+ 6
2+4
3
2
6 µ F and 4 µ F: V ' = × 90 = × 90 =30 V = 90 − 60 V
9
6
Energy stored in each capacitor
1
1
2
3 µ F: CV 2 = × 3 × 10−6 × ( 60 ) = 5.4 mJ
2
2
1
2
6 µ F: × 6 × 10−6 × ( 30 ) =
2.7 mJ
2
1
2
2 µ F: × 2 × 10−6 × ( 60 ) =
3.6 mJ
2
1
2
4 µ F: × 4 × 10−6 × ( 30 ) =
1.8 mJ
2
Example 7. Before the switches are closed, the capacitor C1 has a charge Q1i =C1Vi and C2
has charge Q2i = C2Vi . The initial voltage Vi is common. Find the energy before and after
the switches are closed.
Sol. The total charge is conserved.
( C1 − C2 ) Vi = ( C1 + C2 ) V f
→ Vf =
C1 − C2
Vi
C1 + C2
1
1 ( C1 − C2 ) 2
The final energy
is U f
CeqV f 2
Vi
=
=
2
2 C1 + C2
2
1
The initial energy is U i = ( C1 + C2 ) Vi 2 ≠ U f
2
The difference is dissipated or radiated.
Dielectric Properties
In insulating materials, the permittivity ε is larger than ε 0 . The ratio
κ = ε / ε 0 is called the dielectric constant (dimensionless). In typical
glass, κ is about
2.3. Air has κ 1.00054,
water has κ 80 (dc).
=
=
The origin of higher permittivity in dielectric marerials is due to
polarization of atoms and molecules. In dielectrics, many dipoles
align themselves along the electric field.
If the cpacitor (metal) plate carries a surface charge density σ ,
the surface charge induced on the dielectric surface is
σ induced =
− (1 − κ ) σ
which reduces the electric field to Edielectric
=
(κ − 1) E0
Example 8. Problem 42. Find the dielectric constant of the material inserted.
Ans. If the insulator fills the capacitor, the dielectric constant is
85
=
κ = 3.4 (Nylon)
25
If it fills partially with area A − A ', the capacitance is
A'
A − A'
=
+ κε 0
C ε0
d
d
and the voltage becomes
ε0 A
Q
V= =
C (κ − 1) ε 0 A '+ κ
with A 25
cm 2 and d 1.5 cm is charged to
Example 9. Problem 47. Capacitor
=
=
V = 250V and disconnected from the source. The capacitor is then immersed in water
which has κ = 80. Determine the capacitor (a) charge, (b) voltage, and energy before
and after immersion. (Never do this experiment. Water is not good insulators.)
Sol. The capacitance in air and water are
−4
A
−12 25 × 10
=1.475 × 10−12 F
=
Ca ε 0 =8.85 × 10
−2
1.5 × 10
d
=
Cw κ=
Ca 1.18 × 10−10 F
The charge=
3.69 × 10−10 C remians unchanged because the capacitor is isolated.
Q C=
aV
The voltage in water is reduced by factor κ ,
250
=
Vw = 3.125 V
80
1 Q 2 1 (3.69 × 10−10 ) 2
Energy in air is =
Ua
=
= 4.61 × 10−8 J
−12
2 Ca 2 1.475 × 10
Energy is water is =
U w 5.76 × 10−10 J. The difference in energy is dissipated.
Dipole in Electric Field
Dipole moment (vector) p = qd
where d is the distance between − q and q.
If palced in an electric field E, the dipole is subject to torque
τ = pE sin θ
In vector form, τ =×pE
(N m)
The positive charge is at a potential which is lower than
than the potential of negative charge. Therefore,
the potential energy of a dipole in electric field is
U =− qdE cos θ =−p ⋅ E (J)
E
+q
d
-qE
-q
θ
qE
Example 8. Capacitor partially filled with a dielectric
A fraction f of capacitor (area A, plate distance d ) is filled with dielectric
having dielectric constant κ . Find the capacitance.
A
A
Sol. C1 = ε 0
, C2 = κε 0
fd
(1 − f ) d
=
Ceff
C1C2
=
C1 + C2
κ
ε0 A
f + κ (1 − f ) d
The electric field in the dielectric slab is
smaller than that in air capacitor by
a factor κ . The voltage is divided as
C2
C1
=
V1 =
V , V2
V
C1 + C2
C1 + C2