PHYS2012/2912
ELECTRICAL PROBLEMS
ANSWERS
E039
(a)
For a transparent material, the dielectric constant r can be determined from a
measurement of the refractive index. For a non-magnetic material, n   r .
A dielectric when inserted into a capacitor changes the capacitance value (parallel
plate capacitor C = r0A/d). The capacitance can be determined from a measurement
1
of the resonance frequency of a tuned LC circuit, f 0 
LC
(b)
Polarisability of a molecule
0.5
dilute
dense
0.45
Factor for polarisibility
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
1
1.1
1.2
1.3
ekr
1.4
1.5
(c)
From graph  r, when r > 1.1 the equation for dilute medium starts to fail.
(d)
number density for oxygen gas ng = NA / vol (1 mole contains NA molecules)
vol = 22.410-3 m3
NA = 6.021023
ng = 2.691025 molecules.m-3
number density for oxygen liquid nl =  NA/M
 = 1.19103 kg.m-3 M = 3210-3 kg nl = 2.241028 molecules.m-3
For a dilute medium – oxygen gas  
o
n
( r  1) and   4   o a 3
Equating the two expressions, radius of oxygen atom
  1 
a r
 4 n 
g 

E10_problems_ans.doc
1
3
 1.16 1010 m
a reasonable value
17 aug 10
1
1  rg  1 1  rl  1
1  2K

 rl 
ng  rg  2 nl  rl  2
1 K
 g  l
rl = 1.51
(e)
(f)
where K 
nl  rg  1
ng  rg  2
much greater than the dielectric constant for the gas, because a liquid
has a much larger number density.
p   E   g 
o
ng
( rg  1)
E = 12.5103 V.m-1  = 1.72210-40 F.m2 (C2.N-1.m) p = 2.1510-36 C.m
p
p  q d  8e d d 
 1.68 1018 m much smaller than the diameter of a nucleus
8e
Potential surrounding an electric dipole
1 q q
q  r2  r1 
V ( P) 
  


4   r1 r2  4   r 1r2 
r  d
V ( P) 

q d cos 


d2
4   r 2  cos2  
4


q d cos  p cos 
p r


2
2
4  r
4  r
4  r 3
Potential drops as 1/r2 and so the potential is essentially zero for large r. For a point
charge, the potential drops only as 1/r.
As r  0, then V   hence V is difficult to plot near r = 0 because V is very large.
The plot confirms the above
Conclusions.
E10_problems_ans.doc
17 aug 10
2
E040
When the switch is closed, the capacitor charges and the charge on the capacitor plates at
time t is given by

Q  Qo 1  e
t

  CV 1 e 
t

o
so the charge density on the plates is
Q C Vo
t
f  
1 e 
A
A


where A =  a2 is the area of the plates and the time constant is  = R C. The displacement
current density Jd is equal to the rate of change of the free charge surface density on the
capacitor plates f
Jd 
d f
dt

E10_problems_ans.doc
dD CVo  t   Vo   t 

e 
e
2
dt
A
 a R 
17 aug 10
3
E041
(1)
E
+
-
+
-
+
-
+
shift in atoms
due to ionic nature of bond
induced dipoles due
to shift in electron cloud
rotation
orientation of polar molecules
(2)

e
polarizability of molecule
electronic polarizability
The atomic polarizability is a property of the atom or molecule, it relates to the ease in which
electric dipole moments can be formed giving rise to the polarization of the dielectric
material and hence to the dielectric constant. The value of the polarizability is approximately
the same for solids, liquids and gasses.
(3)
(4)
P  n  Eloc
For xenon gas we can assume that the major contribution to the polarization is due to
the induced dipoles created by the shift in the electron cloud
  e
Eloc  E
n
r  1 
0
pg
N
For a gas pgV  N k T  n  
V kB T
(5)
radius of atom a
  4  0 a 3
(6)
charge separation
d
(7)
(8)
(9)
(10)
electric dipole moment
polarization P  n  E
Pnp
P   r  1  0 E
values ok e.g., radius of atom ~ 10-10 m d is smaller than the nucleus
A helium atom compared with xenon atom
smaller radius  smaller dipole moment  smaller dielectric constant
E10_problems_ans.doc
4  0 a 3
E
Ze
p  Z ed
17 aug 10
4
Using Matlab as a super calculator
clear all; close all; clc
alpha = 3.54e-40
Z = 54
pressure = 5*1.013e5
T = 300
E = 1.45e5
e = 1.602e-19
kB = 1.38e-23
eps0 = 8.85e-12
% alpha
% atomic number
% pressure [Pa]
% temperature
% electric field [V/m]
% electron charge
% Boltzmann constant
% permittivity of free space
n = p / (kB * T)
er = 1 + n*alpha / eps0
a = (alpha / (4*pi*eps0))^(1/3)
d = 4*pi*eps0*a^3*E / (Z*e)
p = Z*e*d
% number density
% dielectric constant
% radius of atom
% charge separation
% electric dipole moment
P1 = n*alpha*E
P2 = (er-1)*eps0*E
P3 = n*p
% polarization
Numerical answers
alpha = 3.5400e-040
Z = 54
pressure = 506500
T = 300
E = 145000
e = 1.6020e-019
k = 1.3800e-023
eps0 = 8.8500e-012
n = 1.2234e+026
er = 1.0049
a = 1.4710e-010
d = 5.9336e-018
p = 5.1330e-035
P1 = 6.2799e-009
P2 = 6.2799e-009
P3 = 6.2799e-009
E10_problems_ans.doc
17 aug 10
5
E073
EQ
+Q
p
-
 0
+
r
Induced dipole
The electric field at the location of the induced dipole due to the point charge is
Q
EQ 
4  r 2
This electric field at the atom induces an electric dipole with electric dipole moment
Q
p   EQ 
4  r 2
The electric field surrounding a dipole is given by
2 p cos 
p sin 
2 p
2 Q
Er 
E 
  180
Er 

E  0
3
3
3
4  r
4  r
4  r  4  2 r 5
2 p
rˆ
4  r 3
The attractive force between the point charge and the induced dipole is
2 Q 2
F  Q Ep 
2
 4   r 5
Ediople 
The force is attractive because the attraction between Q and the induced negative charge is
greater than the repulsion with the induced positive charge.
E095
(a)
When the oil is poured, in the capacitor has already been charged in air and
disconnected from the battery, so the charge on the plates must remain the unchanged.
12
2
 0 A 8.85 10 10 

F  8.85 1011 F
In air: capacitance C 
d
103
Q  CV   8.85  1011  12  C  1.1109 C
Charge
(b)
As the oil is poured the capacitance must increase to  r times its previous value, but
the charge is constant (Q = C V), therefore, the potential difference and the electric
field both must drop by 1/  r times its former value. The electric field is less due to
the presence of the induced charges on the surface of the dielectric.
When the oil is poured in the potential difference remains constant and the
capacitance must increase to  r times its previous value. Therefore, the charge on the
capacitor must increase by the factor  r .
Qoil =  r Qair = (2.2)(1.1×10-9) C = 2.4×10-9 C
The potential difference and hence electric field remain unchanged.
V = Q / C = constant
Q×2.2 & C ×2.2
E = V / d hence E = constant. The induced charge on the surface of the dielectric is
cancelled by the extra free charge on the capacitor plates.
E10_problems_ans.doc
17 aug 10
6
E105
The capacitance is given by C 
Qf
V

f A
V
Applying Gauss’s Law to the top plate E1 
f
f
and to the bottom plate E2 
 r1  0
r2 0
The potential difference between the plates is
   d
d 
V   E dl  E1 d1  E2 d 2   f  1  2 
  0    r1  r 2 
0 A
Hence, the capacitance is C 
d1
d
 2
 r1
r2
E127
Consider a capacitor composed of four plates (M = 4) as shown.
Electric f ield between
Adjacent plates
+q
-q
-q
+q
E
0 A
V  Ed 
+q
-q
q
qd
0 A
Q
V
Q    q    2q   3q
C
C
+V
3 q  0 A 3 0 A

qd
d
In the formula, C = Q/V, Q represents the total charge on either the positive or negative plate.
Hence for M plates, the capacitance is
C
 m  1  0 A
d
and so it behaves like (M-1) capacitors in parallel.
E10_problems_ans.doc
17 aug 10
7
E147
Polarization
P = kr
Bound surface charge density
 b  P nˆ  k R
Bound charge density
 rx ry rz 
 x y z 


  k      3 k
 x y z 
 x y z 
Electric field inside the dielectric sphere
1
1
k r b r
EI   D  P    0  P   

no free charges, D = 0
0
0
 0 3 0
The direction of EI is in the same direction as the vector r .
b   P  k 
Surface charge
QS    b dA   kR   4  R 2   4  k R 3
Interior charge
Qp   b d  (3 k )  4  r 2  dr  4  k R3
R
0
Total charge of dielectric sphere
Q = Qs + Qp = 0
Electric field outside dielectric sphere
EO = 0 since Q = 0.
E10_problems_ans.doc
Gauss’s Law
17 aug 10
 E dA 
qenclosed
0
8
E202
(a)
Gauss’s Law
 D
(b)
D  r 0 E

(c)
(d)
(e)
(f)
 D  f 
dA  q f
E1 
f
 r1  0
E2 
f
r2 0
Q
A
can also consider Gaussian surfaces
  1 
  1 
P1   f  r1  P2   f  r 2 
0
  r1 
 r2 
 Q d  1
1   Q d    r1   r 2 
V   E dl E1 (d / 2)  E2  d / 2   
 



 2  0 A    r1  r 2   2  0 A    r1  r 2 
E
1
C
Q  2  0 A    r1  r 2 



V  d    r1   r 2 
 D  P

P  D  0 E

  1 
  1    f 
  bnet   b1   b 2   f  r1    f  r 2   
   r1   r 2 
  r1 
  r 2    r1  r 2 
Check  r1   r 2  P = 0 
P  b
  1 
  bnet   f  r 2  
 r2 
  1 
  bnet   f  r1  
r2  1
  r1 
  
Also, P  e  0 E   r  1  f 
 r 0 
Alternative derivation of capacitance – assume the two capacitors are in series
1
d
d
C
1/ C1 
1/ C2 
1/ C1  1/ C2
2  r1  0 A
2r2 0 A
 r1  1
 2 A     
C   0   r1 r 2 
 d   r1   r 2 
+Q
 r1
r2
- Qb1
C1
+Q
-Q
+ Qb1
- Qb2
+ Qb2
-Q
+Q
C2
-Q
Capacitors in series
E10_problems_ans.doc
17 aug 10
9
Can find the net bound surface charge density at the interface by applying Gauss’s Law
+Q
- Qb1
E1
+ Qb1
- Qb2
E2
+ Qb2
-Q
 E
dA 
qenclosed
 E1 A  E2 A 
E1 

f
 r1  0
0
1
0
 Qb1  Qb 2 
E2 
f
r2 0
f
f

1

  b1   b 2   net
 r1  0  r 2  0  0
0
 f

   r1   r 2 


 r1 r 2 
 net  
E10_problems_ans.doc
17 aug 10
10
E221
Matlab script
%% E221
close all; clear all, clc;
% data all quantities in SI untis
e = 1.60e-19;
% electron charge
m_e = 9.11e-31;
% electron mass
emf = 12;
% emf of battery
R_int = 5e-3;
% internal resistance of battery
IC = 80;
% cable current
L = 1;
% length of cable
P = 10;
% power dissipated
M = 63.546e-3;
% molecular mass Cu
rho_Cu = 8.9e3;
% density Cu
rho_R = 1.68e-8;
% resistivity Cu
NA = 6.02e23;
% Avogadro's number
% Calculations
V = P / IC;
R = V / IC;
A = rho_R * L / R;
r = sqrt(A / pi);
V_battery = emf - IC * R_int;
JC = IC / A;
n = rho_Cu * NA / M;
v_drift = JC / (n * e);
dt = L /v_drift;
tau = m_e / (n * e^2 * rho_R);
sigma = 1 / rho_R;
lambda = v_drift * tau;
E1 = JC / sigma;
E2 = V / L;
% potential difference across cable
% resistance of cable
% cross-sectional area of cable
% radius of cable
% battery voltage
% current density
% number density
% drift velocity
% time for electron to travel thru cable
% relaxation time
% conductivity
% mean free path
% electric field in cable
% electric field in cable
% Output
fprintf('p.d. across cable, V = %4.2e V\n',V)
fprintf('resistance of cable, R = %4.2e ohms\n',R)
fprintf('cross-sectional area of cable, A = %4.2e m^2\n',A)
fprintf('radius of cable, r = %4.2e m \n',r)
fprintf('battery voltage, V_battery = %4.2e V\n',V_battery)
fprintf('current density, J = %4.2e A/m^2\n',JC)
fprintf('number density, n = %4.2e m^-3\n',n)
fprintf('drift velocity, v_drift = %4.2e m/s\n',v_drift)
fprintf('time for electron to travel thru cable, dt = %4.2e s\n',dt)
fprintf('relaxation time, tau = %4.2e s\n',tau)
fprintf('mean free path, lambda = %4.2e m\n',lambda)
fprintf('conductivity, sigma = %4.2e (ohms.m)^-1\n',sigma)
fprintf('electric field in cable, E1 = %4.2e V/m\n',E1)
fprintf('electric field in cable, E2 = %4.2e V/m\n',E2)
p.d. across cable, V = 1.25e-001 V
resistance of cable, R = 1.56e-003 ohms
cross-sectional area of cable, A = 1.08e-005 m^2
radius of cable, r = 1.85e-003 m
E10_problems_ans.doc
17 aug 10
11
battery voltage, V_battery = 1.16e+001 V
current density, J = 7.44e+006 A/m^2
number density, n = 8.43e+028 m^-3
drift velocity, v_drift = 5.52e-004 m/s
time for electron to travel thru cable, dt = 1.81e+003 s
relaxation time, tau = 2.51e-014 s
mean free path, lambda = 1.39e-017 m
conductivity, sigma = 5.95e+007 (ohms.m)^-1
electric field in cable, E1 = 1.25e-001 V/m
electric field in cable, E2 = 1.25e-001 V/m
Electrons travel very slowly through the cable.
E10_problems_ans.doc
17 aug 10
12
E232
Square plate capacitor A = 0.200 m2
and A = L2
L  A  0.2 m  0.4472 m
x = 0, L/2, L
V = 3.00103 V
Can consider the capacitor as two capacitors in parallel
+ + + + + + + + +
C
r
V
CA
- - - - - - - - -
x
L-x
 0 L x  r  0 L ( L  x)
CB
C = CA + CB
0 L
 x   r ( L  x) 
d
d
d
The charge stored by the capacitor is
  LV 
Q  CV   0
  x   r ( L  x) 
 d 
The energy stored by the capacitor is
  LV 2 
1
U cap  CV 2   0
  x   r ( L  x) 
2
 2d 
As the slab is removed, the change in the energy stored by the capacitor is
  LV 2 
  0 LV 2 
U cap  U cap ( x)  U cap ( x  0)   0
x


(
L

x
)





  r L 
r
 2d 
 2d 
C 


  LV 2 
U cap   0
  x   r ( L  x)   r L 
 2d 
We also have to consider the change in the energy stored by the battery as charge is
transferred from the capacitor to the battery.
Charge transferred to battery = - Charge from capacitor
Charge is conserved qcap + qbattery = 0
qbattery = - qcap
  LV 
  0 LV 
qbattery  (Q( x)  Q( x  0)   0
  x   r ( L  x)   
  r L 
 d 
 d 
  LV 
qbattery    0
  x   r ( L  x)   r L 
 d 
  LV 
  0 LV 
q  (Q( x)  Q( x  0)   0
  x   r ( L  x)   
  r L 
 d 
 d 
  LV 2 
U battery  qV    0
  x   r ( L  x)   r L 
 d 
E10_problems_ans.doc
17 aug 10
13
The work done Wme by the external force Fme is
  0 LV 2 
Wme  U battery  U cap   
  x   r ( L  x)   r L 
 2d 
Differentiating with respect to x gives the force acting on the dielectric
dWme   0 LV 2 
Fme 

  r  1  attractive force between capacitor plates & slab
dx
2
d


which is independent of x and in the direction of increasing x.
x
C
Q
Ucap
Ucap
Ubattery
Wme
Fme
m
10-10 F
10-6 C
10-3 J
10-3 J
10-3 J
10-3 J
10-3 N
x=0
0
5.31
1.59
2.40
0
0
0
3.56
%% E232
close all; clear all; clc;
% data all quantities in SI units unless specified
eps0 = 8.854e-12;
er = 3;
A = 0.2;
d = 10e-3;
V = 3e3;
% Calculations
L = sqrt(A);
x = [0 L/2 L]
C = (eps0 * L / d) .* (x + er*(L-x))
Q = C .* V
U_cap = 0.5 .* C .* V^2
dU_cap = U_cap - U_cap(1)
dU_battery = -(Q-Q(1)) .* V
W = dU_battery + dU_cap
F = (eps0 * L *V^2 / (2*d)) .* (er-1)
E10_problems_ans.doc
x = L/2
0.224
3.54
1.06
1.59
-0.797
1.59
0.797
3.56
x=L
0.447
1.77
0.53
0.797
-1.59
3.19
1.59
3.56
x increasing
Capacitance decreases
Charge on plates decreases
Stored energy decreases
Energy transferred to battery
Work is done on the system
Independent of x
% permittivity of free space
% dielectric constant
% area of plates
% plate separation
% p.d. between plates
% length of capacitor plate
% x positions
% capacitance
% charge
% energy stored by capacitor
17 aug 10
14
E288
0 = 8.8510-12 F.m-1
(a)
A = 20.010-4 m2 d = 4.0010-3 m
Emax_t = 6.00107 V.m-1
Emax_a = 3.00106 V.m-1
r = 2.1
 A
Ca  0  4.4  1012 F
d
The maximum charge depends on the maximum voltage
Vmax_ a  Emax_ a d  1.2 104 V
Qmax_ a  Ca Vmax_ a  5.3 108 C
(b)
Ct   r Ca  9.2 1012 F
Vmax_ t  Emax_ t d  2.4 105 V
(c)
Qmax_ t  Ct Vmax_ t  2.2 106 C
Both the maximum voltage and maximum charge are greatly increased after the Teflon
is inserted.
V = 24.0 V
1
U a  Ca Va 2  1.3 109 J
2
After the Teflon is inserted, C increases by the factor r, whereas V decreases by the
same factor since the charge on the plates remains constant (Q = C V)
2
V  U
1
U t    r Ca     a  6.2 1010 J
2
r
 r 
(d)
Ut < Ua, the potential energy has decreased, the capacitor does work on the Teflon as it
is inserted and so there must be a force that pulls the dielectric in. With sensitive
instruments the tug on the dielectric can be measured.
(e)
The position of the Teflon sheet is immaterial – the field, potential and capacitiance are
independent of the position of the Teflon sheet. By Gauss’s Law, the electric field in
the in the air is
Q
Ea 
 5.6 104 V.m -1
Q = 110-9 C
0 A
The electric field inside the dielectric is reduced by the factor r
E
Et  a  2.7  104 V.m-1
r
C = Q / V  need to find V across the plates


1 
V    E dl  Ea d  Et  d / 2   Ea d 1 
  170 V

 2r 
(f)
C = 610-12 F
This value is intermediate to the capacitance of the system empty or filled with Teflon.
This capacitor is equivalent to two capacitors in series – one empty of width d/2 and the
other filled with Teflon of width d/2.
E10_problems_ans.doc
17 aug 10
15
E293
%% E293
close all; clear all; clc;
% data all quantities in SI units unless specified
T = 300;
% temperature of gas
p = 1.1e5
% pressure of gas
kB = 1.38e-23
% Boltzmann constant
eps0 = 8.854e-12;
% permittivity of free space
a = 5.29e-11;
% Bohr radius - radius of hydrogen atom
e = 1.602e-19;
% electron charge
x = 1e-3;
% plate separation
V = 500;
% p.d. across plates
% Calaculations
n = p/(kB*T);
%number density
alpha = 6.67e-31*(4*pi*eps0);
% atomic polarization
epsR_1 = 1 + n*alpha / eps0;
% dielectric constant 1st estimate
epsR_2 = (2*n*alpha + 3 * eps0) / (3*eps0 - n*alpha); % dielectric constant 2nd estimate
epsR_3 = 1 + 4*pi*n*a^3 ;
% dielectric constant 3rd estimate
E = V/x;
% electric field
d = alpha*V/(e*x);
% separation distance for induced electric dipole moment
R_ad = a/d ;
% ratio a / d
V_i = e*a*x/alpha ;
% Plate voltage – ionization assume dipole distance d = a
W_i = V_i*e*a/x ;
% work done to ionize hydrogen atom [J]
W_ie = W_i/e;
% work done [eV]
% Display answers
disp('Answers')
fprintf('(a) number density, n = %4.2e m^-3\n',n)
fprintf('(b) atomic polarizabiltiy = %4.2e F.m^2 \n',alpha)
fprintf('(c) dielectric constant = %4.5e \n',epsR_1)
fprintf('(d) dielectric constant = %4.5e \n',epsR_2)
fprintf('(e) dielectric constant = %4.5e \n',epsR_3)
fprintf('(f) electric field, E = %4.2e V.m-1 \n',E)
fprintf('(g) dipole: separation distance, d = %4.2e m \n',d)
fprintf('(h) ratio, a/d = %4.2e \n',R_ad)
fprintf('(i) ionization: plate voltage = %4.2e V \n',V_i)
fprintf('(j) ionization: work done, W = %4.2e eV \n',W_ie)
Answers
(a) number density, n = 2.66e+025 m^-3
(b) atomic polarizabiltiy = 7.42e-041 F.m^2
(c) dielectric constant = 1.00022e+000
(d) dielectric constant = 1.00022e+000
(e) dielectric constant = 1.00005e+000
(f) electric field, E = 5.00e+005 V.m-1
(g) dipole: separation distance, d = 2.32e-016 m
(h) ratio, a/d = 2.28e+005
(i) ionization: plate voltage = 1.14e+008 V
(j) ionization: work done, W = 6.04e+000 eV
The answers for the dielectric constant in parts (c) and (d) are the same since the dielectric
constant is very close to 1. The answer (e) is different since the model for the atom used in
the derivation of the equation is crude, however, the model still predicts a value for r close to
1.
E10_problems_ans.doc
17 aug 10
16
The ratio a/d is very large, the separation d of the charges in the induced dipole is minute
even on an atomic scale.
Using our crude model, the work done to ionize the atom (~ 6 eV) which is OK compared
with the 13.6 eV predicted using quantum theory.
E305
dipole moment of water p = 6.210-30 C.m
polarization of water vapour P = ? C.m-2
Pnp
Water vapour gas pressure
pg = 1 atm = 1.013105 Pa
temperature
T = 373 K
Boltzmann constant kB = 1.3810-23 J.K-1
Ideal Gas Law
pg V  N k B T
n
p
N
 g  1.95  1025 molecules.m-3
V kBT
Polarization (maximum – all dipoles aligned with the electric field)
P  n p  1.95  1025  6.2  1030  C.m-2  1.22  104 C.m-2
Langevin function
P
2
np E
3 kB T
p E / k B T  1
P    r  1  0 E
n p2 E
   r  1  0 E
3 kB T
p
3 k B T   r  1  0
n
n
NA
M
kB = 1.3810-23 J.K-1
r = 81
T = 293 K
3
-3
density of water
 = 10 kg.m
molar mass of water M = 1810-3 kg
NA = 6.021023 molecules.mol-1
 0  8.85  1012 F.m-1
n = NA/M =3.351028 molecules.m-3
note: can’t use ideal gas equation to find n
p = 1.610-29 C.m
Water at a lower temperature and as a liquid would have a higher electric dipole moment than
the water vapour. Note: higher number density for liquid compare with gas.
E10_problems_ans.doc
17 aug 10
17
E333
Matlab script
%% E333
close all; clear all, clc;
% data all quantities in SI units unless specified
dx = 60;
% 60 km journey [km]
C_p = 0.1;
% petrol - fuel consumption [L/km]
e_p = 0.20;
% petrol - efficiency
rho_p = 750;
% petrol - density
u_p = 46e6;
% petrol - energy density [J/kg]
u_b = 160;
% battery - energy density [W.h/kg]
u_b = u_b * 3.6e3;
% battery - energy density [J/kg]
e_b = 0.9;
% battery - efficiency
V_C = 240;
% batteries - charging voltage
I_C = 15;
% batteries - charging current
% Calculations
V_p = C_p * dx / 1000;
m_p = rho_p * V_p;
W_p = u_p * m_p;
W_car = e_p * W_p;
% petrol - volume used [m^3]
% petrol - mass used
% petrol - energy supplied
% car - energy as useful work (energy needed for journey)
W_b = W_car / e_b;
m_b = W_b / u_b;
% batteries - energy need to be supplied
% batteries - mass required
P_C = V_C * I_C;
dt = W_b / P_C;
dt = dt / 3.6e3;
% powerpoint - power
% batteries - charging time [s]
% batteries - charging time [h]
% Output
fprintf('petrol - volume used = %4.2e m^3\n',V_p)
fprintf('petrol - mass used = %4.2e kg\n',m_p)
fprintf('petrol - energy supplied = %4.2e J\n',W_p)
fprintf('car - energy as useful work = %4.2e J \n',W_car)
fprintf('batteries - energy need to be supplied = %4.2e J\n',W_b)
fprintf('batteries - mass required = %4.2e kg\n',m_b)
fprintf('powerpoint - power = %4.2e W\n',P_C)
fprintf('batteries - charging time = %4.2e h\n',dt)
petrol - volume used = 6.00e-003 m^3
petrol - mass used = 4.50e+000 kg
petrol - energy supplied = 2.07e+008 J
car - energy as useful work = 4.14e+007 J
batteries - energy need to be supplied = 4.60e+007 J
batteries - mass required = 7.99e+001 kg
powerpoint - power = 3.60e+003 W
batteries - charging time = 3.55e+000 h
The required mass of the lithium-ion batteries is about 80 kg, which is quite reasonable. The
time to charge the batteries from a normal powerpoint is only about 4 h. Hence, for short
journey such as this, an electric car powered by lithium-ion batteries is very feasible.
E10_problems_ans.doc
17 aug 10
18
E361
(a)
1
U cap  CV 2  7.2 103 F
2
C = 10010-6 F V = 12 V
(b)
Ubattery  QV  4.3106 J
Q = 100 A.h = (100)(3.6103) A.s = 3.6105 C
(c)
2U battery
1
U battery  CV 2 V 
 3 105 V very large value
2
C
(d)
1
U battery  CV 2
2
(e)
time constant  = R C = 1.510-4 s discharge time t = 5  = 7.510-4 s
Charge on capacitor plates decreases exponentially
C
t
(f)
2U battery
V
2
 6 104 F very large value
t
Q  Qo e RC  CV e RC  CV e5  8 106 C
In this time energy delivered by battery
V2
U battery 
t  7 102 J
R
Maximum current from battery
V
I max   8 A
battery current can be much greater than 8 A
R
E10_problems_ans.doc
17 aug 10
19
E382
The electric field in the air part Eair – apply Gauss’s Law where the Gaussian surface is a
cylinder of radius r, length (L-h) and the for the inner tube, the linear charge density is air


 D dA  qenclosed D  2  r  L  h     L  h  D  2 airr Eair  2  air 0 r
The electric field in the air part Eoil – apply Gauss’s Law where the Gaussian surface is a
cylinder of radius r, length h and the for the inner tube, the linear charge density is oil


 D dA  qenclosed D  2  r  h  oil h D  2 oilr Eoil  2  oilr  0 r
The potential difference of the inner tube to the outer tube is
a
a 
  dr 
  
  
In the air
V   E dl     air       air  ln  a / b    air  ln(b / a)
b
b
 2  0   r 
 2  0 
 2  0 
In the oil
a
a
 oil 
 oil 
oil   dr 
V    E dl    

ln  a / b   


 ln(b / a)


b
b
 2  r  0   r 
 2  r  0 
 2  r  0 
From the two expressions for V
oil   r air
The total charge on the inner cylinder is
Q   L  h  air  h oil  air  L   r  1 h 
The capacitance of the two metal tubes is
 2   0    2   0   L    r  1 h  
Q
C   air  L    r  1 h  



V
ln(b / a)
 air ln(b / a)  

The capacitor formed by the two cylinder is connected to the battery, therefore, as the oil is
attracted into the space between the cylinders the capacitance increases and so the charge on
the tubes must increase (C = Q / V V = constant & C   Q  ). Therefore, charge is
transferred from the battery and its energy decreases and the stored energy of the capacitor
increases
Usystem = Ucapacitor + Ubattery
dUsystem/dh = dUcapacitor/dh + dUbattery/dh
dU system d  1
d 1
dC
 d
 d
  CV 2    QV    CV 2    CV 2   V 2
dh
dh  2
dh  2
dh
 dh
 dh
Therefore, the magnitude of the force Foil attracting the oil into the tubes is
dU system  V 2  dC
Foil 
 
dh
 2  dh


  0   r  1V 2
dC  2   0  r  1
Now
 Foil 

ln(b / a)
dh
ln(b / a)
The magnitude of the gravitational force FG opposing the movement of the oil into the
capacitor is
FG  m g     b 2  a 2  h g
Equating the two forces and rearranging the height to which the oil rises is
 0   r  1 V 2
h
  b 2  a 2  g ln(b / a )
E10_problems_ans.doc
17 aug 10
20
E456
Gauss’s Law
(a)
(b)
(c)
(d)
(e)
 D
dA  q f from the spherical symmetry D is constant for any radius r
Q
radial outward
4 r 2
When a charged object (+Q) is immersed in a dielectric, the electric field surrounding
the object decreases. This is because of the polarization of the charges in the dielectric
at the surface of the charged object (bound surface charge Qb) and the consequent
screening of the charges on the object by the polarization field within the dielectric.
Q  Qb
D
Q
E


radial outward
2
 r  0 4  r  0r
4  0r 2
From the two expressions for the electric field
  1 
Qb  Q  r

 r 
The electric, displacement and polarization fields are related by
1
Q   r 1 
E   D  P  hence the polarization is P 

 radial outward
0
4 r 2   r 
 D
q f  Q
dA  D (4  r 2 )  Q
D
The bound charge density b is related to the divergence of the polarization
  r 1  3
 Q    r  1   rˆ 
 Q    r 1 
3
  2   
  4   r    Q 
 r 


 4    r   r 
 4    r 
 r 
b   P   
Bound charge at centre of dielectric material is found by integrating the bound charge
density over a spherical volume
 
(f)
b
  1 
  1 
  1 
d   Q  r   3 (r ) d  Q  r    3 (r ) d  Q  r 
 r 
 r 
 r 
The bound charge surface density b is related to the polarization by
  P nˆ where n̂ is the normal unit vector pointing away from the dielectric at the
surface.
Q   r  1 
At r = a (inside surface)  ba  P nˆ 


4 a2   r 
At r = b (outside surface)
 bb  P nˆ 
Q   r  1 


4  b2   r 
N.B. direction of P and n at the inner and outer surfaces
and  ba   bb
(g)
since a < b
Total charge of dielectric material
Qdielectric  Qba  Qbb   ba (4 a 2 )   bb (4 b 2 )
The dielectric remain neutral.
  1 
Qba  Q  r

 r 
Qdielectric  0
E10_problems_ans.doc
17 aug 10
  1 
Qbb  Q  r

 r 
21
The electric field lines and charge
distribution – free and bound charges
E
+
+
The electric outside the dielectric is
identical to that of a point charge +Q. The
dielectric remains neutral and by Gauss’s
Law the electric field is only determined by
the net charge enclosed by a surface. The
sudden change in the number of electric
field lines within the dielectric is due to the
polarization of the dielectric. The dielectric
does not shield the charge on the metal
sphere in the region outside the spherical
dielectric layer.
E10_problems_ans.doc
17 aug 10
+
+
+
- + +-
+ +
++
+
E
+ = 0+
++
+
++++
- -
+
+
+
22
E551
Can consider the fuel gauge capacitor as two capacitors in parallel.
Ceq 
 eff  0 l w
d
Ceq  Cair  C fuel
Cair 
 0 l  h w
C fuel 
d
r 0 h w
d
h
 
 eff  1      r  1
l
Matlab
h/L
0.00
0.25
0.50
0.75
1.00
35
30
25
eff
20
e
%% E551
close all; clear all; clc;
% data all quantities in SI units unless specified
er_p = 1.95;
er_m = 33;
h = [0 0.25 0.5 0.75 1];
% ratio h/L
% Calculations
e_p = 1 + h .* (er_p - 1);
e_m = 1 + h .* (er_m - 1);
% Output
disp('petrol methanol h/L e_eff')
for c = 1 : 5
fprintf('%6.2f % 6.2f %6.2f \n',h(c),e_p(c),e_m(c))
end
% Plot
hp = linspace(0,1,200);
ep_p = 1 + hp .* (er_p - 1);
ep_m = 1 + hp .* (er_m - 1);
figure(1)
plot(hp,ep_p,'linewidth',2)
hold on
plot(hp,ep_m,'linewidth',2)
xlabel('h/L')
ylabel('e_{eff}')
grid on
15
10
5
0
0
0.1
0.2
0.3
0.4
0.5
h/L
0.6
0.7
0.8
0.9
1
petrol methanol
e_eff
1.00 1.00
1.24 9.00
1.48 17.00
1.71 25.00
1.95 33.00
The variation in eff is much greater for methanol than petrol, hence the gauge for methanol is
more practical.
E10_problems_ans.doc
17 aug 10
23
E567
f = 3.00 C.m-2 = 3.0010-6 C.m-2
Free charge surface density
Dielectric constant
r = 3.00
0 = 8.85410-12 C2.N-1.m-2
Outside the dielectric
D =  f = 3.00 C = 3.0010-6 C.m-2
EO 
f
 3.39 105 V.m-1
0
PO = 0
Inside the dielectric
D =  f = 3.00 C = 3.0010-6 C.m-2
E
1
EI   D  PI   O =1.13105 V.m-1
0
r
PI  D   0 EI  2.00 106 C.m-2
 b  PI nˆ  2.00 106 C.m-2
Alternatively
E
1
 D  PI   O
0
r

1
 f b   f

0
r 0


 b   f 1 
1  2
   2.00  106 C.m -2
 r  3 f
If D is represented by 12 field lines then
E0 by 12 field lines
EI by 12/3 = 4 field lines
+ + + + + + + + + + + +
- + +
- -
+
+
+
+
+
+
- - - - - - - - - -
PI by (12-8) = 8 field lines
 f
b
 b
 f
 f
b
 b
 f
Electric displacement
E10_problems_ans.doc
Electric f ield
17 aug 10
Polarization
24
E582
density of sulfur
atomic number
molar mass
internal electric field
dielectric constant
(1 to 3)
 = 2.1x103 kg.m-3
Z = 16
M = 32x10-3 kg
E = 107 V.m-1
r = 4
Induced dipole moment – sulfur atom
E
-8e
+16e
-8e
Zero electric field –
helium atom
symmetric  zero
dipole moment
+16e
-8e
A
-8e
d
B
effectively charge +16e at A and -16e at B
dipole moment p = 16 e d
p
(4)
(5)
(6)
The electric field polarizes the sample of
P  e  0 E  n p
r  4
e   r  1  3
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
+
-
+

msample
V
n
 NA
p
 0 e E
(7)
M
n
Nm
M

n
V
NA
+
 2.1  10  6.02  10  atoms.m
3

(8)
p  Z ed
+
+
+
+
+
 b
 b
23
3

sulfur
E
32  10
8.85  1012   3 1 107 
3.95  10
28
-3
 3.95  1028 atoms.m-3
C.m  6.722  1033 C.m
6.722 1033 

p
d

m  2.6 1015 m
19
Z e (16)(1.602 10 )
(9)
The separation distance d is much smaller than the diameter of the atom
d ~ nuclear dimensions
 NA
P  e  0 E  n p  n Z e d
e   r  1
n
M
(10)
   1 M E
d 0 r
Z e  NA
E10_problems_ans.doc
17 aug 10
25
E588
DC corresponds to a fixed polarity for the voltage e.g., when a dielectric is placed between
the plates of a capacitor connected to a battery. Optical corresponds to the frequencies
associated with visible (light) electromagnetic radiation
Optical frequencies ~ 51014 Hz
For transparent, non-magnetic materials, the refractive index of the material depends on the
dielectric constant, therefore, the refractive index is frequency dependent n   r .
Virtually, there is no change in the dielectric constant of diamond because the induced
dipoles produced by electron cloud displacements can occur almost instantaneously and can
follow the applied electric field even at high frequencies.
Both NaCl and LiCl are ionic materials and the dielectric constant is mainly due to ion
displacements. The ion displacements can’t follow the high frequency alternating electric
field and so the dielectric constant drops as the frequency is increased.
Water is a polar molecule and the dielectric constant depends on the orientation of the
molecules. The big drop in the dielectric constant is because the molecules can’t rotate fast
enough to align with the applied electric field and so at high frequencies the dielectric
constant is much smaller.
E600
No dielectric C 
0 A
d
8.85 10 110  F  8.85 10
12

2
110
3
11
F
Lines of D are continuous across perpendicular boundaries, so D is the same in both the air
gap and dielectric sheet. To find the capacitance C = q / V, find expressions for q and V.
Gauss’s Law D   f 
q
A
qDA
V    E dl Eair (d  t )  Edielectric t 
C
D
0
(d  t ) 
 D 
t 
   d  t  
r 0  0 
r 
Dt
0 A
q

V 
t 
d t  
r 

8.85 1012 1102  

  r 1  
0 A   1  
3
 m  6.2 105 m
t 
 d 
    110 
10


C
2
1.01

10
  
 r 

E10_problems_ans.doc
17 aug 10
26
E617
+Q/2
V1
C1
+Q/2
- Q/2
- Q/2
+q A
V2
C1 = Q / 2V1
C1
+q B
C1
C2
- qA
Check: if r = 1  V2 = V1
Q = 2 C 1 V1
q A = C 2 V2 =  r C 1 V2
q B = C 1 V2
Q = qA + qB
= C1 V2 (r + 1)
= 2 C 1 V1
V2 = 2 V1 / (r + 1)
- qB
q A = 2 C 1 V1 r / (r + 1)
q B = 2 C 1 V1 / (r + 1)
qA = qB = Q/2
E664
Condition (1): the potential difference between the plates of the capacitor remains constant
(V = constant) as the plates are moved further apart (y increases). As the plates are moved
A
further apart the capacitance must decrease ( C 
), therefore, the charge on the plates
y
must decrease (Q = CV) as well as the electric field (by Gauss’s Law). Since the charge and
electric field decrease with increasing separation of the plates, the force between the plates
must also decrease.
Condition (2): The charge on the plates remains constant (Q = constant) and hence electric
field is also constant (E = constant). So the force between the plates maintains its initial value
during the entire motion as the plates separate.
Therefore, for the same displacement of the plates, the work done under condition (2) is
greater than under condition (1).
E10_problems_ans.doc
17 aug 10
27
E696
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
Gaussian surface 1
+Qf
top
+ + + + + + + + + + + + + + + + +
(d-t)/2
-Qb
-
-
-
-
-
-
d
t
+Qb
+
+
+
+
+
+
+
(d-t)/2
-Qf
- - - - - - - - - - - - - - - - - - - - bottom
Gaussian surface 2
the plates since E  V. The free charges Qf on the conductive plates of the capacitor remains
unchanged.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
- +
- +
- +
- +
- +
-
 f
b b
+
+ - +
+
+
+ - +
+
+
+ - +
+
+
+ - +
+
+
+ - +
+
E
E10_problems_ans.doc
P
17 aug 10
o
- 
f
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
- +
- +
- +
- +
- +
D
-
o
28
%% E696
close all; clear all; clc;
% data all quantities in SI units unless specified
eps0 = 8.854e-12; % permittivity of free space
er = 3;
% dielectric constant
A = 0.2;
% area of plates
d = 10e-3;
% plate separation
V = 3e3;
% p.d. between plates
t = 5e-3;
% thickness of dielectric
% Calculations
C_o = eps0 * A / d;
% capacitance - no dielectric
E_o = V / d;
% electric field - no dielectric
Q_of = C_o * V;
% free charge on plates - no dielectric
sigma_of = Q_of / A; % free charge surface density - no dielectric
D_o = sigma_of;
% electric displacement - no dielectric
U_o = 0.5 * C_o * V^2; % energy stored by capacitor - no dielectric
vol = A * d;
% volume of space between capacitor plates
u_o = U_o / vol;
% energy density - no dielectric
E_do = V/(d-t+t/er);
% electric field - dielectric - free space
E_dd = E_do / er;
% electric field - dielectric - in dielctric
Q_df = eps0 * E_do * A; % free charge on plates - dielectric
C_d = Q_df / V;
% capacitance - dielectric
sigma_df = Q_df / A; % free charge surface density - dielectric
D_d = sigma_df;
% electric displacement - no dielectric
sigma_db = sigma_df*(1-1/er); % bound charge surface density - dielectric
P = sigma_db;
% polarization
U_d = 0.5 * C_d * V^2; % energy stored by capacitor - dielectric
u_d = U_d / vol;
% energy density - dielectric
dU_cap = U_d - U_o % change in energy stored by capacitor
dU_battery = (Q_of - Q_df) * V % chnage in energy stored by battery
% Output
disp('NO DIELECTRIC')
fprintf('capacitance, C = %4.2e F\n',C_o)
fprintf('electric field, E = %4.2e V/m\n',E_o)
fprintf('free charge on plates, Q = %4.2e C\n',Q_of)
fprintf('free surface charge density, sigma_f = %4.2e C/m^2\n',sigma_of)
fprintf('electric displacement, D = %4.2e C/m^2\n',D_o)
fprintf('energy stored by capacitor, U = %4.2e J \n',U_o)
fprintf('energy density, u = %4.2e J/m^3\n',u_o)
disp('DIELECTRIC')
fprintf('capacitance, C = %4.2e F\n',C_d)
fprintf('electric field (free field), E = %4.2e V/m\n',E_do)
fprintf('electric field (inside), E = %4.2e V/m\n',E_dd)
fprintf('free charge on plates, Q = %4.2e C\n',Q_df)
fprintf('free surface charge density, sigma_f = %4.2e C/m^2\n',sigma_df)
fprintf('electric displacement, D = %4.2e C/m^2\n',D_d)
fprintf('bound surface charge density, sigma_b = %4.2e C/m^2\n',sigma_db)
fprintf('polarization, P = %4.2e C/m^2\n',sigma_db)
fprintf('energy stored by capacitor, U = %4.2e J \n',U_d)
fprintf('energy density, u = %4.2e J/m^3\n',u_d)
fprintf('change in energy stored by capacitor, dUcap = %4.2e J \n',dU_cap)
fprintf('change in energy stored by battery, dUbattery = %4.2e J \n',dU_battery)
E10_problems_ans.doc
17 aug 10
29
With dielectric in place
Edielectric  E free space /  r
Voltage between the plates:


t 
V    E dl  E free space  d  t   Edielectric t  E free space  d  t  

r 

Electric field between plates
E free space
V
E free space 
Edielectric 
r

t 
d t  
r 

Gauss’s Law - Gaussian surface applied to (+) plate and adjacent free space region
Q
E free space A  f
Q f   0 E free space A
0
Capacitance
Q
Cdielectric  f
V
Gauss’s Law - Gaussian surface applied to (+) plate and dielectric
E free space

1
1
Edielectric 
 f   f   b    D  P 
r

 b   f 1 

r 0
0
0
1

r 
+Qf
-Qf
E10_problems_ans.doc
17 aug 10
30
E704
(a)
Assume vacuum filled capacitor to start.
Radius of inner conducting sphere (charge +Q), R1
Radius of outer conducting sphere (charge –Q), R2
Q
C
capacitance
V
R1
2
potential
V    E.dl
Gauss’s Law
 E.dA 
electric field between spheres
E
1
R2
qenclosed
0
Q
4 0 r2
(b)
potential from electric field
R2 dr
Q
Q 1 1 
Q R2  R1
V  |V |

 

2

R
4 0 1 r
4   0  R1 R2  4   0 R1 R2
(c)
Q 4   0 R1 R2
C 
capacitance
V
R2  R1
Capacitance depends only on the geometry
(d)
When dielectric added between conducting spheres, can replace 0 by the permittivity of the
material  = r 0.
4   r  0 R1 R2
C
capacitance with dielectric
R2  R1
Capacitance increases by the factor r (r > 1) because the presence of the dielectric decreases
the electric field between the plates (E smaller, V smaller, Q constant, C larger)
(e)
For large radii of curvature and as R2  R1
R1 R2 = R2 and R2 – R1 = d A = 4 R2
  A
C r 0

capacitance of a parallel plate capacitor
d
(f)
Isolated conducting sphere capacitor
4 
two spheres
C
1
1

R1 R2
Take limits R1  R and R2  
capacitance of isolated sphere C  4   R where  is the permittivity of medium
surrounding the sphere.
(g)
1
 9 109 N.C-2 .m 2 RE  6.38 106 m
Capacitance of the Earth    0 k 
4 0
CEarth = 7.110-4 F
E10_problems_ans.doc
17 aug 10
31
E741
(a)
C
U
...
...
...
V
Q
V
1 Q2 1
1
 QV  CV 2
2 C 2
2
Capacitors in series (charge on each plate)
Ctotal 
Series
branch
1
1
1

 ...
C1 C2
Capacitors in parallel (voltage across
each capacitor is the same)
Ctotal  C1  C2  ...
Matlab script
%% E741
close all; clear all; clc;
% data all quantities in SI units unless specified
U = 4.60e7;
% energy required for journey
V = 100;
% voltage across capacitor bank
C_u = 3000;
% individual capacitance
V_u = 2.0;
% individual voltage
% Calculations
% Total capacitance
Q = 2 * U / V;
C = Q / V;
% total charge stored
% total capacitance
% Individual capacitor
Q_u = C_u * V_u;
U_u = 0.5 * Q_u * V_u;
N_u = U / U_u;
% charge on individual capacitor
% energy stored by individual capacitor
% number of individual capacitors
% Series branch
N_s = V / V_u;
Q_s = Q_u;
C_s = Q_u / V;
C_sc = C_u / N_s;
U_s = 0.5 * Q_s * V;
% number of capacitors in each series branch
% charge stored in each series branch
% capacitance of series branch
% capacitance of series branch CHECK
% energy stored in series branch - same as an individual capacitor
% Parallel branches
N_p = N_u / N_s;
Cc = N_p * C_sc;
N_pc1 = U / U_s;
N_pc2 = Q / Q_u;
% number of series branches
% total capacitance CHECK
% number of series branches CHECK
% number of series branches CHECK
E10_problems_ans.doc
17 aug 10
32
% Output
fprintf('total charge, Q = %4.2e C\n',Q)
fprintf('total capacitance, C = %4.2e F\n',C)
fprintf('charge stored on each capacitor, Q_u = %4.2e C\n',Q_u)
fprintf('energy stored by each capacitor, U_u = %4.2e J\n',U_u)
fprintf('number of individual capacitors, N_u = %4.2e \n',N_u)
fprintf('number of capacitors in each series branch, N_s = %4.2e \n',N_s)
fprintf('charge stored in each series branch, Q_s = %4.2e C\n',Q_s)
fprintf('capacitance of series branch = %4.2e C\n',C_s)
fprintf('energy stored in series branch, = %4.2e J\n',U_s)
fprintf('number of series branches = %4.2e \n',N_p)
Results
total charge, Q = 9.20e+005 C
total capacitance, C = 9.20e+003 F
charge stored on each capacitor, Q_u = 6.00e+003 C
energy stored by each capacitor, U_u = 6.00e+003 J
number of individual capacitors, N_u = 7.67e+003
number of capacitors in each series branch, N_s = 5.00e+001
charge stored in each series branch, Q_s = 6.00e+003 C
capacitance of series branch = 6.00e+001 C
energy stored in series branch, = 3.00e+005 J
number of series branches = 1.53e+002
The number of capacitors required for this short trip is quite large, more than 7000.
E761
Consider case (1) capacitor with no dielectric inserted and case (2) dielectric inserted into
capacitor. In both cases the capacitor is connected to the battery which holds the potential
different constant  V = constant V = V1 = V2.
0 A
r 0 A
C2 / C1   r since only the dielectric has changed the
d
d
capacitance has increased by the factor  r .
Capacitance C1 
C2 
Charge Q1  C1 V Q2  C2 V   r C1 V   r Q1 since only the dielectric has changed the
capacitance has increased by the factor  r , the charge has also increased by the factor  r .
Electric field (parallel plate capacitor) E = V / d neither V or d has changed, therefore, E is
the same.
1
1
1
Stored energy U1  C1 V 2 U 2  C2 V 2   r C1 V 2   rU1 stored energy increases by
2
2
2
the factor  r .
The dielectric is attracted to the charged plates of the capacitor and is drawn into the region
between the plates and so work is done on the dielectric slab. This work done on the slab is
equal to the work done by the battery (negative) as charge is transferred to the capacitor plus
the increase in stored energy of the battery (positive).
E10_problems_ans.doc
17 aug 10
33
E797
(a)
The electric displacement is only determined by the free charges and is uniform throughout
the space between the capacitor plates. D   f
(b)
In the free space, the electric field E is determined by the free charge. E 
f
0
The effect of the dielectric is to reduce the electric field inside it by the factor (1/r)

2 f
E1  f
E2 
20
3 0
(c) The polarization is determined from


2 f  f
1
E   D  P
P1   f  f  f   b1
P2   f 

  b2
0
2
2
3
3

1 2  25 a  f

(d)
V    E dl  E (3a)  E1 (a)  E2 (a)  a  f  3    
2 3
6


(e)
(f)
E 0
 f
E
S1
 f / 2
E1
S3
 f / 2
E
 f / 3
E2
S2
 f / 3
E
S4
 f
E 0
Gaussian surfaces and Gauss’s Law
q
qenclosed (free & bound charges)
 E dA  enclosed
0
f  f
1
 f 

0 
2  20
  
1 
S2: E1    f  f   f
0 
2  20
  2 f
1
S3: E2    f  f  
0 
3  3 0
S1: E1 
S4: E2 
 f  2 f
1 
  f 

0 
3  3 0
E10_problems_ans.doc
17 aug 10
values agree with part (b)
34
E815
(a) The capacitance only depends upon the geometry and not the charge on the plates of the
capacitor
C
Q 0 A

V
d
(b)
Apply Gauss’s Law to the –Q plate with a close cylindrical Gaussian surface, then,
q
Q
 E
 E dA  enclosed
0
0 A
By symmetry the electric field must be uniform between the plates of the capacitor.
Applying a Gaussian surface to the positive plate and Gauss’s Law, the charge in the
inner surface must of the positive plate must be +Q. The electric field inside a
conductor is zero, therefore a charge of +0.2Q must be located on the outer surface.
(c)
Potential difference between the plates is


Qd
V   E.dl  E  dl  E d 
0 A


(d)
Interior points electric
+0.2Q on outer surface
field must be zero
+
+
+ + + + + + + + + +
+Q on inner surface
-
- - - - - - - - Interior points electric
field must be zero
E851
Resistance
R
d
A
-Q on inner surface
Symmetry
– electric field must be uniform
– electric field lines perpendicular to conductive plates
Capacitance
Potential difference across the plates
C
C
r 0 A
d
RC    r  0
Q
Q
V
V
C
The conduction current is due to a flow of electrons through the dielectric from the positive
plate to the negative plate
V
Q
Q
iC  

R R C  r 0
The displacement current is due to the change in the charge on the capacitor plates. Assume
the charges decreases exponentially
dQ 1
Q
Q  Qo e t / RC
iD 

Qo e t / RC 
dt RC
 r 0
The net current is iC + iD = 0
E10_problems_ans.doc
iC  iD
17 aug 10
35
E888
(a)
3 0  r  1
n r  2
2 -1

polarizability [C .N .m]– ease in which electric dipoles are induced in molecules in
an external electric field p   E
   e e electronic polarizability – induced electric dipole moment due to shift in electron
cloud
n
number density [molecules.m-3] – number of electric dipole per volume
permittivity of free space  0  8.85  1012 F.m-1
0
Claussius-Mossotti equation

dielectric constant (relative permittivity) of medium  r  1
The Claussius-Mossotti equation relates the microscopic parameter, the polarizibility  of a
molecule to the macroscopic measureable quantity, the dielectric constant r for materials
containing non-polar molecules.

n
r  1 r  2  3
  0   r  1
r  1
(b)
n
0
-40
2 -1
(c)
 = 2.335510 C .N .m
r
(1)
radius of He atom, a
1/ 3
  
10
  4  0 a
a
  1.3 10 m
 4 0 
4
p = ? C.m E = 10 V.m-1
p   Eloc   E  2.34  1036 C.m
3
(2)
(3)
correct order of magnitude
p = Qd
Q = 2e (for He) d = ? m
e = 1.60210-19 C
d = p / 2e = 7.310-18 m
smaller than the diameter of a nucleus
3
4  0 a
E
Could also use d 
Ze
 = 2.335510-40 C2.N-1.m  0  8.85  1012 F.m-1
n
r  1
 need to find n
0
gas pressure
pg = 1 atm = 1.013105 Pa
temperature
T = 293 K
Boltzmann constant kB = 1.3810-23 J.K-1
Ideal Gas Law
p
N
pg V  N k B T
n   g  2.51  1025 molecules.m-3
V kBT
n
r  1 
 1.00066
Dilute gas
0
can’t round to 1 r -1 is used to determine the number if significant figures not r
Always re-arrange an equation first with unknown on LHS then substitute numbers
In calculating and using an equation do not include un-necessary vectors indictors,
e.g. P  n p which is not sensible
(4)
r = ?
E10_problems_ans.doc
17 aug 10
36
E890
 0  8.85  1012 F.m-1
Q = 510-6 C
(a)
A = L2 = 0.25 m2
L = 0.5 m
d = 110-3 m
r = 2.3
surface charge induced on the surface of the dielectric (Gauss’s Law)

Eair
Q  Qb
Q
1
Eair 
Edielctric 

Qb  Q 1  
0 A
r 0 A 0 A
 r 
Qb = 2.810-6 C
Edielctric 
Q
= 9.8105 V.m-1
(b)
electric field
(c)
potential difference V = E d = 9.8102 V
(d)
capacitance C 
(e)
1
1
stored energy U  QV  CV 2 = 2.510-3 J
2
2
r 0 A
d
r 0 A

Q
 5.110-9 F
V
E893
 b  P n so at one end of the cylinder b = + P and at the other end b= - P as the unit
vector n points away from the dielectric. There are no are no bound charges in the interior of
the dielectric since P is uniform b   P  0
E10_problems_ans.doc
17 aug 10
37
E896
Gaussian surface S
(1)
conducting
sphere q
air
a
r
non-conducting
liquid
Symmetry  field lines must be radial
(2) (3) (4)
 D
dA  qenclosed
S
 D
air
Sair

dA 
Dliquid dA  Dair  2 r 2   Dliquid  2 r 2   q
Sliquid
Dair  Dliquid 
q
 2 r 
2
(5)
The relationship between the electric and displacement fields is given by D   E
Dair   0 Eair
Diquid   r  0 Eliquid
(6)
radial electric field
conducting
sphere q
air
Eairt
Eliquidt
non-conducting
liquid
Symmetry 
(7)
from parts (4) and (5)
0 E  r 0E 
(8)
Eairt = Eliquidt  Eair = Eliquid = E
q
 2 r 
2
E
q
2  0   r  1 r 2
from parts (5) and (7)
Dair 
E10_problems_ans.doc
q
2  r  1 r 2
17 aug 10
Dliquid 
r q
2  r  1 r 2
38
(9)
(10)
On the surface of the conducting sphere r = a
r q
q
 air  Dair (a) 
 liquid  Dliquid (a) 
2
2  r  1 a
2  r  1 a 2
Greater concentration of free charge on the bottom of the sphere compared to top.
Greater charge on bottom  increase in E, dielectric  decrease in E  effects
cancel – E uniform value around conducting sphere at any given radius r > a.
r = 1
E
q
2  0  r  1 r
2

q
q

2
2  0 1  1 r
4  0 r 2
which is the electric field surrounding a point charge  ok
q
q
q


2
2
2  r  1 a
2 1  1 a
4 a 2
which is the charge divided by the surface area of the sphere  ok

(11)
Sphere is a conductor – electric field inside must be zero. The charge q is located on
the outer surface of the sphere.
field lines of D
(12)
+
+
+
+
+
+
+
+
+ +
greater concentration of charge
on surface bounded by liquid
(13)
field lines of E
(14)
field lines of D
Polarization of the dielectric in the liquid near the surface of the conductor produces
bound charges which tends to reduce the electric field in the liquid but this is exactly
compensated by the increase in the free charges located on the surface of the
conductor in the liquid  same electric field pattern surrounding the charged sphere
exposed to the air and liquid.
E10_problems_ans.doc
17 aug 10
39
E900
By symmetry the electric field points in an outward radial direction. The total free charge on
4

the sphere is Q f   f   a 3  . The electric displacement and electric field are related by
e

the equation E 
(a)
D
r 0
Apply Gauss’s Law for D
Electric field inside the sphere (r < a)
 D
f r
4

 D(4  r 2 )   f   r 3   D 
3
3

dA  qenclosed
S
E
f r
r
3 r  0
Electric field outside the sphere (r > a)
 D
dA  qenclosed
S
E
(b)
 f a3
4 3
 D(4  r )   f   a   D 
3r2
3

2
 f a3
r
3 0 r 2
Potential V at the centre of the sphere ( R   V  0 )
Since the field points radially outwards, we integrate along a radius from infinity to
zero. The differential element dl points out from the origin, so E dl  E dr
0
a
0


a
V  V (0)  V ()    E dl    Eoutside dr   Einside dr
a  a
V    b 2
 3 r
 0
3
 
V   b
 3 0
(c)
0  r

b
 dr   a 

 3 r  0

 dr

0
a
   a3 
 b
 r2  
      
    
   r    2 r  a 
 3 0
  2 a2 
  a 

2 r 

  a2  
1 
V   b  1 

 3  0   2 r 
The polarization P of the dielectric is related to the electric field E = Einside inside the
dielectric. The polarization and electric field both have a direction which is radial
outward.
  r    r    1 
P   0 e E   0  r  1  b    b   r 
 3 0  r   3    r 
Therefore, at the surface of the dielectric the bound surface charge density is
  a    1 
 b  P( r  a )   b   r 
 3  r 
And the total bound surface charge is
E10_problems_ans.doc
17 aug 10
40
  f a    r 1 
4
   r 1 
2
3
qb ( surface)   b  4  a 2   
 4 a     a  f  


3
 r 
 3  r 
(d)
  1 
qb ( surface)  Q  r

 r 
In the interior of the dielectric sphere, the bound charge density is related to the
polarization
 f
 3
b   P   P  r  
   r  1   x y z 
 f
     

   r   x y z 
 3
  r 1 
   r 1 
  3    f 


 r 
 r 
And the total interior bound charge is
  1   4
  1 
4


qb (interior)   b   a3     f  r    a3   Q  r 
3


 r  3
 r 
(e)
(f)
From parts (c) and (d), the total bound charge is zero
qb(surface) + qb(interior) = 0
1
The energy density is u   r  0 E 2 and to find the total energy U of the system will
2
need to integrate the energy density over all space taking into account the spherical
symmetry

d  

0
 4  r  dr
2
1
u  r 0 E2
2
1
1
u   r  0 Einside 2   0 Eoutside 2
2
2
uinside
  r 
1
 r 0  f 
2
 3 0  r 
uinside
 f2

 18  
0 r


d  

0
2
 2
 r

uoutside
uoutside
3
1  f a 
 0 

2  3  0 r 2 
2
  f 2 a6   1 

 18    r 4 
0 

 4  r  dr
2
2
2 6
a  4  f 
  4  f a   1 
4
4
U  
r
dr



  2  r dr

0  18   
a 
0 r 

 18  0   r 
 4   f 2   a5   4   f 2 a 6   1 
U 

  
 18     5   18 
 
0 r 
0

 a 
 2   f 2 a5   1  5  r 
U 
 

 45 
0

 r 
E10_problems_ans.doc
17 aug 10
41
E906
(a) (b)
E
Q
0 A

V    E.dl   E  d  t    0  t   E  d  t  

(c)
C
Q
d  t 
0 A
 A
Q
 0
V (d  t )
(d)
Symmetry
– fields must be uniform
– field lines perpendicular to plates
Interior points electric
field must be zero
+ + + + + + + + + +
+Q on inner surface
- - - - - - - - - + + + + + + + + + +
-Q on outer surface
+Q on outer surface
-
-Q on inner surface
- - - - - - - - -
E
dl
Interior points electric
field must be zero
E10_problems_ans.doc
17 aug 10
42