Electrycity 2 p c r o 1

AZAD SLEMAN HSENAZAD SLEMAN HSEN
ELECTRYCITY AND
MAGNETIC
ELECTRYCITY AND
MAGNETIC
COLLEGE OF SCINCECOLLEGE OF SCINCE
physic departmentphysic department
20112011

• Electric field
• Electric potential
• Capcitor
• Current
• Resistance
• Electric field
• Electric potential
• Capcitor
• Current
• Resistance

‫ئه‬ ‫دكوريت‬ ‫خو‬ ‫جهي‬ ‫باركه‬ ‫كو‬ ‫مي‬ ‫ده‬
‫يدادبن‬ ‫به‬ ‫فه‬ ‫ئه‬ ‫دكوريت‬ ‫خو‬ ‫جهي‬ ‫باركه‬ ‫كو‬ ‫مي‬ ‫ده‬
‫يدادبن‬ ‫به‬ ‫فه‬
POSITIVE charge moving in an E field.
(a) When a positive charge moves in the direction of an
electric field, the field does positive work and the
potential energy decreases. Work = qo E d

The charge qo moves along a straight line
extending radially from charge q.

Electric field E
and potential V
at points inside
and outside a
positively
charged
spherical
conductor.
See
www.physics.sjsu.edu/becker/physics51

Coulomb’s law:
EpW
dqEW
.
.
=
=
Electric field:
Work = Force x distance
2
04
1
r
q
E
πε
=
r
q
rV
04
1
)(
πε
=
WdFPEdqEPEPE ==∆=−=− ..12
r
QQ
rU
21
4
1
)(
0πε
=
 SI Unit of electric potential: Volt (V)
electric potential=potential Energy/charge
dE
q
dqE
V .
.
==
JMNW ,.=

2/122
00 )(4
1
4
1
dx
dx
r
dq
dV
dxdq
+
==
=
λ
πεπε
λ
 A rod of length L located along the x
axis has a uniform linear charge
density λ. Find the electric potential at
a point P located on the y axis a
distance d from the origin.
( )[ ]
( )[ ]ddLL
dxx
dxx
dx
dx
dVV
L
LL
ln)(ln
4
)(ln
)(
2
1
.2ln
4)(4
2/122
0
0
2/122
0
2/1122
00
2/122
0
−++=
++












++=
+
== −
∫∫
πε
λ
πε
λ
πε
λ





 ++
=
d
dLL
V
2/122
0
)(
ln
4πε
λ
xdr ′′+′′=
azad

P
R r
x
x
Every dQ of charge on the ring is the
same distance from the point P.
2 2
dq dq
dV k k
r x R
= =
+
2 2ring ring
dq
V dV k
x R
= =
+
∫ ∫ 2 2
kQ
V
x R
=
+
B) Use Ex = - dV/dx Ex = - (x2
+ a2
)
dV -1/2
kQx
dx
= - kQ
d
dx
Ex =
= - kQ (- ½)
(x2
+ a2
)
3/2
2x(x2
+ a2
)
-3/2
What about Ey and Ez ?
What is the electric potential at the center
of the ring?
What is the electric field at the center of
the ring?

Example: A disc of radius R has a uniform charge per unit area σ and total
charge Q. Calculate V at a point P along the central axis of the disc at a
distance x from its center.
P
r
x
xR
The disc is made of concentric rings. The
area of a ring at a radius r is 2πrdr, and the
charge on each ring is σ(2πrdr).
We *can use the equation for the
potential due to a ring, replace R
by r, and integrate from r=0 to
r=R.
R
2 2 2 2ring ring 0
0 0
1 2 rdr rdr
V dV
4 2x r x r
σ π σ
= = =
πε ε+ +
∫ ∫ ∫
( ) ( )
R
2 2 2 2 2 2
2
0 0 00
Q
V x r x R x x R x
2 2 2 R
σ σ
= + = + − = + −
ε ε πε
2
Q
R
σ =
π
( )2 2
2
0
Q
V x R x
2 R
= + −
πε
rx ′′+′′
dQ
Remembe (r+x).r its uniform with
intgral in content mrs azad
Note:in the z direction you change z place
X ok and you can countinio on the exampl:

(B) Find the electric potential at point P located on the
perpendicular central axis of a uniformly charged disk of
radius a and surface charge density σ. (b) Find an
expression for the magnitude of the electric field at point P.
(b) Ex = - dV/dx
2πkσ (x2
+ a2
)
1/2
- x
Ex = -
dV
dx
= -
d
dx
x2
+ a2
x
1 -Ex = 2πkσ
x2
+ a2
x
1 -2πkσ
When you are really close to this disk, then it is as if you are
looking at an infinite plane of charge, use above equation to
deduce the electric field. Is the result consistent with the result
obtained from our discussion using Gauss’s law ?

Electric Potential due to uniformly charged annulus
Calculate the electric potential at point P on the axis
of an annulus, which has uniform charge density σ.
drrdAdq
xr
dqk
dV e
πσσ 2where
22
==
+
=
V= 2πσk [ (x2
+b2
)1/2
- (x2
+a2
)1/2
]
=
+
= ∫
b
a
e
xr
drr
kV
22
2 σπ
Ok you can restory
this function

Electric Potential due to non-uniformly charged disk
A disk of radius R has a non-uniform surface charge density
σ=Cr where C is constant and r is distance from the center
of the disk as shown. Find the potential at P.
dV =
ke dq
r2
+ x2
dq = σdA = Cr(2πrdr) and
∫ +
=
R
e
xr
drr
kCV
0
22
2
)2( π
Standard
integral
V= C(2πk) {R(R2
+x2
)1/2
+x2
ln(x/[R+ (R2
+x2
)1/2
])}
r

Electric Potential due to a finite line of charge
A rod of length 2 l located along the x axis has a total charge Q
and a uniform linear charge density λ=Q/2l . Find the electric
potential at a point P located on the y axis a distance a from the
origin.
l
V = kλ
dxl
x2
+ a2
- l
V =
kQ
2 l
ln
l 2
+a2
+ l
l 2
+a2
- l
How is this result consistent with the E field for infinite
line of charge obtained using Gauss’s Law?

Electric Potential due to a finite line of charge
A rod of length L as shown lies along the x axis with its left end
at the origin and has a non-uniform charge density λ = αx (where
α is a positive constant). (a) What are the units of α?
(b) Calculate the electric potential at point A.
(c) Calculate the electric potential at point B.
(a) C/m2
V = k ∫ r
dq
= k ∫ r
dxλ =
+
=
∫
L
xd
dxx
k
0
)(
α
= kα [ L – d ln( 1 + L/d) ]

Electric Potential due to a uniformly charged sphere
An insulating solid sphere of radius R has a uniform
positive volume charge density and total charge Q.
(a) Find the electric potential at a point outside the sphere,
that is, r > R. Take the potential to be zero at r = .
(b) Find the potential of a point inside the sphere, (r < R).
Outside the sphere, we
have
Er =
k Q
r2
For r > R
To obtain potential at B, we use
VB = -
r

Er dr = - kQ
r
r2
dr
VB =
k Q
r
Potential must be continuous at r = R, => potential at
surface
VC =
k Q
R

Inside the sphere, we have
k Q
Er =
R3
For r < R
To obtain the potential difference at D,
we use
VD - VC = -
r
Er dr = - r dr =
r
R
k Q
R3
r
R
k Q
2R3
( R2
– r2
)
VC =
k Q
R
Since
To obtain the absolute value of the potential at D, we
add
the potential at C to the potential difference VD - VC :
VD =
k Q
2R
3 -
r2
R2
Check V for r = R
For r < R

Outside the sphere, we
have
k Q
Er =
r2
For r > R
To obtain potential at B, we
use
VB = -
r

Er dr = - kQ
r
r2
dr
VB =
k Q
r

Electric Potential due to a uniformly charged sphere
Pictures from Serway & Beichner
Inside the sphere, we have
k Q
Er =
R3
For r < R
To obtain the potential difference at D, we
use
VD - VC = -
r
Er dr = - r dr =
r
R
k Q
R3
r
R
k Q
2R3
( R2
– r2
)
VC =
k Q
R
Since
To obtain the absolute value of the potential at D, we add
the potential at C to the potential difference VD - VC :
VD =
k Q
2R
3 -
r2
R2
Check V for r =
R
For r < R

Example: Two connected Charged Conducting Spheres
Two spherical conductors of radii r1 and
r2 are separated by a distance much
greater than the radius of either sphere.
The spheres are connected by a
conducting wire. The charges on the
spheres in equilibrium (the spheres are
at the SAME electric potential V) are q1
and q2 respectively, the they are
uniformly charged. Find the ratio of the
magnitudes of the electric fields at the
surfaces of the spheres.
Connected  both must have the same electric potential.
V = kq1/r1 = kq2/r2  q1/r1= q2/r2
Sphere far apart  charge uniform  magnitude of surface E given
by
E1 = kq1/r1
2
and E2 = kq2/r2
2
we have E1/E2 = r2/r1

L
P
a
0
a
La +
x
dxλdQ =
x
Q
kV
d
d =
∫∫
+
==
La
a x
x
kdVV
dλ
La
a
xk
+
= lnλ
( )[ ]aLak lnln −+= λ
a
La +
ln
x
 A rod of length L located along the X axis has a uniform linear
charge density λ. Find the electric potential at a point P located
on the X axis a distance a from the origin.
x
dx
kV
λ
=d

ring
Example 25 Q41 here
Consider a ring of radius R with the total charge Q spread
uniformly over its perimeter. What is the potential difference
between the point at the center of the ring and a point on its
axis a distance 2R from the center?
( )
R
Qk
R
Qk
R
Qk
RR
Qk
VVV
e
e
ee
R
553.0
1
5
1
2
22
02
−=






−=
−
+
=−=∆
P
R r
x
Q
2R

Electric Potential due to a charged
conductor
(a) The excess charge on a
conducting sphere of radius R is
uniformly distributed on its
surface.
(b) Electric potential versus distance r
from the center of the charged
conducting sphere.
(c) Electric field magnitude versus
distance r from the center of the
charged conducting sphere.
Pictures from Serway & Beichner

Example: a 1 µC point charge is located at the origin and a -4 µC point charge
4 meters along the +x axis. Calculate the electric potential at a point P, 3
meters along the +y axis.
q2q1
3 m
P
4 m
y
i 1 2
P
i i 1 2
-6 -6
9
3
q q q
V = k = k +
r r r
1×10 -4×10
= 9×10 +
3 5
= - 4.2×10 V
 
 ÷
 
 
 ÷
 
∑

Example: find the total potential energy of the system of three charges.
q2q1
3 m
P
4 m
x
y
q3
1 2 1 3 2 3
12 13 23
q q q q q q
U = k + +
r r r
 
 ÷
 
( )( ) ( )( ) ( )( )-6 -6 -6 -6 -6 -6
9
1×10 -4×10 1×10 3×10 -4×10 3×10
U = 9 10 + +
4 3 5
 
 ÷×
 ÷
 
-2
U = - 2.16 10 J×

Example (Motion in Uniform Field(
An insulating rod having linear charge density λ = 40.0 µC/m
and linear mass density µ = 0.100 kg/m is released from rest
in a uniform electric field E = 100 V/m directed perpendicular
to the rod as shown. (a) Determine the speed of the rod after
it has traveled 2.00 m. (b) How does you answer to part (a)
change if the electric field is not perpendicular to the rod ?
(Ignore gravity)
Arbitrarily take V = 0 at the initial point, so at
distance d downfield, where L is the rod length
V = − Ed and Ue = − λLEd
(K + U)i = (K + U)f
0 + 0 = 1
2 µLv 2
− λLEd
v =
2λEd
µ
2(40.0 × 10
−6
C/m)(100 N/C)(2.00 m)
(0.100 kg/m)
= 0.4 m/s
=

Potential Energy
Three point charges are
fixed at the positions
shown. The potential
energy of this system of
charges is given by
U = k
q1 q2
r12
q1 q3
r13
q2 q3
r23
+ +

Example: Electric Potential due to a dipole
An electric dipole consists of two charges
of equal magnitude and opposite sign
separated by a distance 2a. The dipole is
along the x-axis and is centered at the
origin. (a) Calculate the electric potential
at point P. (b) Calculate V and Ex at a
point far from the dipole. (c) Calculate V
and Ex if P is located anywhere between
the two charges.
V = qi
riΣk
i
= k
q
x- a x+ a
q
=
2kqa
x2
- a2
(a)
(b) x >> a, V ~ 2kqa/x2
Ex =- dV/dx = 4kqa/x3
(c)
V =
qi
ri
Σk
i
= k
q
a - x x+ a
q
=
2kqx
x2
- a2
Ex =
dV
dx
=
d
dx
2kqx
x2
- a2
= 2kq
-x2
-a2
x2
- a2 2

Electric Potential Energy: a point charge moves from i to f
in an electric field, the change in electric potential energy
is
WUUU if −=−=∆
Electric Potential Difference between two points i and f
in an electric field:
q
U
q
U
q
U
VVV if
if
∆
=−=−=∆
Equipotential surface: the points on it all have the same
electric potential. No work is done while moving charge on it.
The electric field is always directed perpendicularly to
corresponding equipotential surfaces.
r
q
rV
04
1
)(
πε
=
Potential due to point charges: ∫ ⋅−=
∆
≡∆
f
i
sdE
q
U
V

0
Potential due to a collection of point charges:
∑∑ ==
==
n
i i
i
n
i
i
r
q
VV
101 4
1
πε
Potential of a charged conductor is constant everywhere inside
the conductor and equal to its value to its value at the surface.
∫∫ ==
r
dq
dVV
04
1
πε
Calculatiing E from V: s
V
Es
∂
∂
−=
z
V
Ez
∂
∂
−=
x
V
Ex
∂
∂
−=
y
V
Ey
∂
∂
−=
Electric potential energy of system of point charges: r
qq
VqU 21
0
2
4
1
πε
==
summary

CapacitorCapacitor
a device that stores chargea device that stores charge
made of two conductors separated by an insulator
The amount of charge that a
capacitor can store depends on:
1. area of conducting surface
2. distance between the conductors
3. type of insulating material
link

CapacitanceCapacitance
the ratio of chargecharge to potential differencepotential difference
C = Q/VC = Q/V Q = CVQ = CV
The SI unit of capacitance
is the FaradFarad, FF, named in
honor of
Michael FaradayMichael Faraday
FaradayFaraday.
One Farad of capacitance means that
one Coulomb of charge may be stored
in the capacitor for each Volt of
potential difference applied.

Capacitor CircuitsCapacitor Circuits
SeriesSeries
1. reciprocal of the1. reciprocal of the total capacitancetotal capacitance is the sumis the sum
of the reciprocals of the separate capacitorsof the reciprocals of the separate capacitors
1/C1/CTT = 1/C= 1/C11 + 1/C+ 1/C22 + 1/C+ 1/C33 + ...+ ...
2.2. chargecharge is the same on each capacitoris the same on each capacitor
QQTT = Q= Q11 = Q= Q22 = Q= Q33 = ...= ...
3.3. total potential differencetotal potential difference is the sum of eachis the sum of each
VVTT = V= V11 + V+ V22 + V+ V33 + ...+ ...
In other words, in a series circuit,In other words, in a series circuit,
capacitance adds as reciprocals,capacitance adds as reciprocals,
charge stays the same,charge stays the same,
and voltage adds.and voltage adds.

CC11
CC22
CC33
EE = 12 V= 12 V
CCTT = 4.0= 4.0 µµ FF
VVTT = 12 V= 12 V
QQTT = 48= 48 µµ CC
CC11
CC22
CC33
C,C,
µµ FF
V,V,
VV
Q,Q,
µµ CC
12
10
15
48
48
48
4.0
4.8
3.2
48124
1Ct
=×
=∆× QV

ParallelParallel
1.1. total capacitancetotal capacitance is the sum of eachis the sum of each
separate capacitorseparate capacitor
CCTT = C= C11 + C+ C22 + C+ C33 + ...+ ...
2.2. total chargetotal charge is the sum of the chargesis the sum of the charges
on each separate capacitoron each separate capacitor
QQTT = Q= Q11 + Q+ Q22 + Q+ Q33 + ...+ ...
3.3. potential differencepotential difference is the same acrossis the same across
each capacitoreach capacitor
VVTT = V= V11 = V= V22 = V= V33 = ...= ...
In other words, in a parallel circuit,In other words, in a parallel circuit,
capacitance and charge add,capacitance and charge add,
but voltage stays the same.but voltage stays the same.

CC33
EE = 12 V= 12 V
CCTT = 22= 22 µµ FF
VVTT = 12 V= 12 V
QQTT = 264= 264 µµ CC
CC11
CC22
CC33
C,C,
µµ FF
V,V,
VV
Q,Q,
µµ CC
8
10
4
CC22
CC11
12
12
12
96
120
48
•CCeqeqVV == CC11VV+C+C22V+V+CC33VV
• 264264 ==8.128.12+10.12++10.12+4.124.12

Series CapacitorsSeries Capacitors
VVtotaltotal= V= V11+V+V22+V+V33
VCVCeqeq= V= V11CC11= V= V22CC22= V= V33CC33 = Q= Q
1/C1/Ceqeq= 1/C= 1/C11+1/C+1/C22+1/C+1/C33
-- ++
+Q+Q
+Q+Q
+Q+Q
-Q-Q
-Q-Q
-Q-Q
Parallel CapacitorsParallel Capacitors
•QQtotaltotal= Q= Q11+Q+Q22+Q+Q33
•CCeqeqV = CV = C11V+CV+C22V+CV+C33VV
CCeqeq= C= C11+C+C22+C+C33
--
++
CC11 CC22 CC33
VV
QQtotaltotal QQ11 QQ22 QQ33
Q = CVQ = CV

Created by: azad sleman university of duhok 29
Capacitance
Capacitors in Parallel and Series:
Equivalent capacitance:
Example: (a) Find the equivalent capacitance of the combination as shown. Assume C1=12.0 µF,
C2=5.30 µF, C3=4.50 µF. (b) A potential difference V=12.5 V is applied to the input terminals.
What is the charge on C1?
Solution: (a):
(b):
Example: A 3.55 µF capacitor C1 is charged to a potential difference V0=6.30 V, using a 6.30 V
battery. The battery is then removed and the capacitor is connected as in the figure to an
uncharged 8.95 µF capacitor C2. When switch S is closed, charge flows from C1 to C2 until the
capacitors have the same potential difference V. What is the common potential difference?
Solution:
(series)
11
(parallel)
11
∑∑ ==
==
n
i ieq
n
i
ieq
CC
CC
312123
2112
111
CCC
CCC
+=
+=
F57.3
F3.17
123
12
µ
µ
=
=
C
C
V58.2C6.44
12
123
12
12
12123123 =====
C
q
C
q
VVCq µ
VCVCVC
qqq
2101
210
+=
+=
V79.1
21
1
0 =
+
=
CC
C
VV

Capacitor & capacistanc
We usually talk about capacitors in terms of
parallel conducting plates
They in fact can be any two conducting
objects
abV
Q
C =
Units are
Volt
Coulomb
farad
1
1
1 =
The capacitance is defined to be the ratio of
the amount of charge that is on the capacitor
to the potential difference between the plates
at this point
The magnitude of the electric field between the
two plates is given by A
Q
E
00 εε
σ
==
We treat the field as being uniform allowing
us to write A
dQ
EdVab
0ε
==
d
A
C oε=
‫ره‬ ‫دناف‬ ‫نه‬
‫دي‬ ‫م‬ ‫ئه‬ ‫بتيدا‬
‫قانوني‬ ‫يييي‬‫بف‬
‫ين‬ ‫جيكه‬

fffarad
d
A
C o
0
µµµ
ε
35401054.31054.3
105/21085.8
39
312
=×=×=
=×××==
−−
−−
coulombsVCq
ab
649
1054.3101054.3
−−
×=××==
coulombsVCq
ab
649
1054.3101054.3
−−
×=××==
mCAq /1077.12/1054.3/
56 −−
×=×==σ
mvE /1021077.11036/
659
0
×=×××==
−
πεσ
mvdVE
ab
/102105/10/
634
×=×==
−
mCED /1077.11021085.8
5612
0
−−
×=×××==ε

Cylindrical Capacitor: (Optional)
The charge resides on the outer surface of the inner conductor and the
inner wall of the outer conductor. Assume the length of the cylinder L>>b.
( )
( )ab
L
C
a
b
L
Q
dr
rL
Q
EdrV
b
a
b
a
ln
2
ln
22
0
00
πε
πεπε
=






=== ∫∫
Spherical Capacitor: (Optional)
( ) ab
ab
C
ba
Q
dr
r
Q
EdrV
b
a
b
a −
=





−=== ∫∫ 0
0
2
0
4
11
44
πε
πεπε

Example: An isolated conducting sphere whose radius R is 6.85 cm has
a charge q=1.25 nC. (a) How much potential energy is stored in the
electric field of this charged conductor? (b) What is the energy density at
the surface of the sphere? (c) What is the radius R0 of an imaginary
spherical surface such that half of the stored potential energy lies within
it?
Solution:
( ) ( )
( ) ( ) ( ) RR
U
RR
q
drr
r
q
drruc
R
q
Eubn
R
q
C
q
Ua
R
R
R
R
2
2
11
8
4
4
1
2
1
4:
J/m4.25
4
1
2
1
2
1
:J103
82
:
0
00
2
2
2
2
0
0
2
3
2
2
0
0
2
0
0
22
00
==





−=





=
=





=====
∫∫ πε
π
πε
επ
µ
πε
εε
πε
Potential Energy and Energy Density:
2
2
2
1
2
CV
C
Q
U ==
2
0
2
1
Eu ε=
The electric potential energy of a
capacitor is the energy stored in the
electric field between the two plates
(electrodes). It is the work required to
charge the capacitor
The energy density is the
potential energy per unit
volume

1-In the figure, battery B supplies 12 V. (a) Find the
charge on each capacitor first when only switch S1
is closed and (b) later when switch S2 is also closed.
Take C1=1.0 µF, C2=2.0 µF, C3=3.0 µF, and C4=4.0
µF.
Home work
2-A parallel-plate capacitor of plate
area A is filled with two dielectrics
as shown in the figure. Show that
the capacitance is
2
210 εεε +
=
d
A
C
3-A parallel-plate capacitor of plate
area A is filled with two dielectrics of
the same thickness as shown in the
figure. Show that the capacitance is
21
2102
εε
εεε
+
=
d
A
C

Calculating the Capacitance
Putting this all together, we have for the capacitance
d
A
V
Q
C
ab
0ε==
The capacitance is only dependent upon the geometry of the capacitor
Example
A circuit consists of three unequal capacitors C1, C2, and C3 which are connected
to a battery of emf Ε. The capacitors obtain charges Q1 Q2, Q3, and have voltages
across their plates V1, V2, and V3. Ceq is the equivalent capacitance of the circuit.
a) Q1= Q2 b) Q2= Q3 c) V2= V3 d)
E = V1 e) V1 < V2 f) Ceq > C1
A detailed worksheet is available detailing the answers
The answer it is a green coulor

Redefinitions
We now redefine several quantities using the dielectric constant
d
A
d
A
KKCC εε === 00
with the last two relationships holding for a parallel plate capacitor
Energy Density
22
0
2
1
2
1
EEKu εε ==
We define the permittivity of the dielectric as being
0εε K=
Capacitance:

Energy Stored in a Capacitor
Electrical Potential energy is stored in a capacitor
The energy comes from the work that is done in charging the
capacitor
Let q and v be the intermediate charge and potential on the
capacitor
The incremental work done in bringing an incremental charge,
dq, to the capacitor is then given by
C
dqq
dqvdW
ab
==
Force between two plate capacitor drFdW .=
Energy density
ADF .2/
2
ε=AdxDdW
AxDW
.2/
.2/
2
2
ε
ε
=
=
ε2/
2
D=

16.9 Energy Stored in a Capacitor
Average voltage during charging:Average voltage during charging:
SinceSince VVfinalfinal is the applied voltage, we writeis the applied voltage, we write VVaa==VV/2./2.
Energy stored (=work done by the battery):Energy stored (=work done by the battery):
22
finalinitialfinal
a
VVV
V =
−
=
ADpEdqEdfvqcqEW
CVqVqVW
.2/...2/12/
2
1
2
1
222
2
a
ε=∆======
===
0
The stored energy is: W = 0.5CV2
= 0.5 x .
005 x1002
= 25 J

Example
d
A
- - - - -
+ + + +
Now suppose you pull the plates further apart so
that the final separation is d1
Which of the quantities Q, C, V, U, E change?
How do these quantities change?
Answers: C
d
d
C
1
1 = V
d
d
V 1
1 =
Suppose the capacitor shown here is charged to Q
and then the battery is disconnected
U
d
d
U 1
1 =
Charge on the capacitor does not change
Capacitance Decreases
Voltage Increases
Potential Energy Increases
Q:
C:
V:
U:
E: Electric Field does not change

Suppose the battery (V) is kept
attached to the capacitor
Again pull the plates apart from d to d1
Now which quantities, if any, change?
How much do these quantities change?
Answers:
Example
C
d
d
C
1
1 = E
d
d
E
1
1 =U
d
d
U
1
1 =
Q:
C:
V:
U:
E:
Voltage on capacitor does not change
Capacitance Decreases
Charge Decreases
Potential Energy Decreases
Electric Field Decreases
Q
d
d
Q
1
1 =

Electric Field Energy Density
The potential energy that is stored in the capacitor can be thought of as
being stored in the electric field that is in the region between the two
plates of the capacitor
The quantity that is of interest is in fact the energy density
dA
uDensityEnergy
VC 2
2
1
==
where A and d are the area of the capacitor plates and their separation,
respectively
Using
d
A
C 0ε= and dEV = we then have
2
0
2
1
Eu ε=
Even though we used the relationship for a parallel capacitor, this
result holds for all capacitors regardless of configuration
This represents the energy density of the electric field in general

Dielectrics
Most capacitors have a nonconducting material between their plates
This nonconducting material, a dielectric, accomplishes three things
1) Solves mechanical problem of keeping the plates separated
2) Increases the maximum potential difference allowed between the plates
3) Increases the capacitance of a given capacitor over what it
would be without the dielectric
Suppose we have a capacitor of value C0 that is charged to a potential
difference of V0 and then removed from the charging source
We would then find that it has a charge of 00VCQ =
We now insert the dielectric material into the capacitor
We find that the potential difference decreases by a factor K K
V
V 0
=
Or equivalently the capacitance has increased by a factor of K
0CKC =
This constant K is known as the dielectric constant and is
dependent upon the material used and is a number greater than 1

16.10 Capacitors with Dielectrics
• A dielectric is an insulating material that, when placed
between the plates of a capacitor, increases the capacitance
– Dielectrics include rubber, plastic, or waxed paper
• C = κCo = κεo(A/d)
– The capacitance is multiplied by the factor κ when the
dielectric completely fills the region between the plates
Capacitance in presence of a dielectric:
d
A
C
C
V
Q
/V
Q
V
Q
C
0
0
0
000
ε=
====
κ
κ
κ
κ
Since κ>1, the dielectric
enhances the capacitance
of the capacitor!

(a) Electric field lines inside an
empty capacitor
(b) The electric field produces
polarization
(c) The resulting positive and
negative surface charges on
the dielectric reduce the
electric field within the
dielectric
E0
E=E0/κ or V=V0/κ
+Q0-Q0
Reasoning:
Dielectric constant

the value of κ DepenDs on the
nature of the Dielectric material,
as the table below inDicates

Capacitors Designs
(a)Paper capacitor
(b)High-voltage oil capacitor
(c) Electrolytic capacitor

Two identical parallel plate capacitors are given the same charge Q, after
which they are disconnected from the battery. After C2 has been charged
and disconnected it is filled with a dielectric.
Compare the voltages of the two capacitors.
a) V1 > V2 b)V1 = V2 c) V1 < V2
Example
We have that Q1 = Q2 and that C2 = KC1
We also have that C = Q/V or V= Q/C
Then
1
1
1
C
Q
V = and 1
1
1
2
2
2
1
V
KKC
Q
C
Q
V ===

Electric Current
The electric current is the amount of
charge per unit time that passes
through a plane that pass completely
through the conductor.
The SI unit for current is a coulomb per second (C/s), called as an
ampere (A)
Avnq
t
Q
I
d
=
∆
∆
= nqAIv
a
/=

Microscopic Description of Current: Math
τ
m
Enq
J
2
=
•Assume uniform motion and density of charge carriers
A
•The charge in a wire of length L can be calculated
L
q=(nAL)ev, for electrons
•The total charge moves through a cross-section in:
t=L/v ; v is the drift velocity
A
•Assume uniform motion and density of charge carriers
•I=q/t=nALev/L =nAev
•This implies (J=I/A) that J=(ne)v
ne is the charge carrier density

Conductivity
•In most materials, electric field is required to make the current
move
•the current is proportional to the electric field, and the
conductivity σ
•The conductivity is a property of the
material
EJ σ=
Empirically,
EJ =ρ
•the resistivity is the reciprocal of the conductivity, nothing
more!
Current I
Electric Field E

Direction of current
• A current arrow is drawn in
the direction in which positive
charge carriers would move,
even if the actual charge
carriers are negative and
move in the opposite
direction.
• The direction of conventional
current is always from a point
of higher potential toward a
point of lower potential—that
is, from the positive toward
the negative terminal.

A complete circuit is one where current can flow
all the way around. Note that the schematic
drawing doesn’t look much like the physical
circuit!
Electric current

current density J
If the current is constant and perpendicular to a surface, then and
we can write an expression for the magnitude of the current
density
J=
i
A nevJ
eqnqvJ
−=
−=⇔=
Daft speed

Current Density
When a conductor is in electrostatic equilibrium, its conduction electrons move randomly
with no net direction. In the presence of electric field, these electrons still move randomly,
but now tend to drift with a drift speed, in the direction opposite to that of the applied
electric field that causes the current. The drift speed is tiny compared to the speed in the
random direction. For copper conductors, typical drift speeds are from 10-5
to 10-4
m/s as
compare to 106
m/s for random motion speeds
For convenience, the drift of positive charged
particles is in the direction of the applied electric
field E. Assume all particles move at the same drift
speed and the current density is uniform across the
wire’s cross sectional A. The number of particles in
a length L of the wire is nAL, where n is the number
of particles per unit volume. Therefore the total
charged particles in the length L, each with charge e,
is
q=(nAL)e
Since the particles all move along the wire with
speed vd this total charge q moves though any cross-
section of the wire in the time interval
t = L / vd In vector form

Suppose, instead that the current density through a cross-
section of the wire varies with radial distance r as J = ar2
which a=3 x 1011
A/m2
and r is in meters. In this case what
is the current through the outer portion of the wire
between radial distance R/2 and R ?
Since the current density is NOT uniform,
We have to resort to the integral relationship
I = ∫ J • dA and integrate the current density over
the portion of the wire from r = R/2 to R
The current density vector J (along the wires length) and
the differential area vector dA (perpendicular to the cross
section of the wire) have the same direction. Therefore
We need to replace the differential area dA with
something we can integrate between the limits r=R/2
and r=R. The simplest replacement is the area 2πrdr of
a thin ring of circumference 2πr and thickness dr. We
can then integrate with r as the variable
Problem

Problem
From the previous slide, we must solve the integral

67
Example - current through a wire
• The current density in a cylindrical wire of radius R=2.0 mm is uniform
across a cross section of the wire and has the value 2.0 105
A/m2
. What
is the current i through the outer portion of the wire between radial
distances R/2 and R?
• J = current per unit area = di / dA
RArea A’ (outer portion)
Current through A’

The resistance
• The resistance (R) is defined as the ratio
of the voltage V applied across a piece of
material to the current I through the
material: R=V/i.
.SI Unit of Resistance: volt/ampere
(V/A)=ohm(Ω)

69
Resistance vs Resistivity
• The conducting properties of a material are
characterized in terms of its resistivity.
• We define the resistivity, ρ, of a material by the ratio
E: magnitude of the applied field
J: magnitude of the current
density
ρ =
E
J
V
m




A
m2




=
Vm
A
= Ωm
R = ρ
L
A
For a wire
ρ is measured in Ω m
L – length of the conductor
A – its area.
• The units of resistivity are
resistivity = resistance x cross section area/ length:

Example III–2. Problem 17.17 (Page 590) from the Serway
& Voile textbook: A wire 50.0 m long and 2.00 mm in diameter
is connected to a source with a potential difference of 9.11 V,
and the current is found to be 36.0 A. Assume a temperature of
20C and, using Table 17.1, identify the metal of the wire.
Example about Resistivity
Ω==
∆
= 253.0
36
12.9
A
V
I
v
R
m
m
m
L
dR
L
RA
.1059.1
)50(4
)102()253.0()4/(
8
232
Ω×
×Ω
===
−
−
ππ
ρ
It is a silver

25 x 103 Ω
±10%
RESESTANE AND
RESISTOR
‫بكارهينانا‬ ‫بي‬ ‫ي‬ ‫بكه‬ ‫مقاومي‬ ‫هزمارا‬ ‫دشي‬ ‫تو‬ ‫ريكي‬ ‫بفي‬
‫بينه‬ ‫ر‬ ‫بجه‬ ‫جهازي‬

73
Resistance and Temperature
• When electrons move through the conductor they
collide with atoms:
– Resistivity grows with temperature ( more collisions)
ρ0 – resistivity measured at some reference temperature T0
α – temperature coefficient of resistivity
When electrons move through the conductor they collide with atoms:
Temperature of the conductor increases because of the current
(through collisions)
Electrical energy is transformed into thermal energy
Resistors dissipate energy
Power – energy per unit of time- (in W=J/s) dissipated by a resistor
Ρ ιs the resistivity at some temperature T.

)](1[
00
TTaRR −+=
If a wire is of constant cross-sectional area A and length L with
respect to change in temperature, we can write
Example III–3. At 40.0C, the resistance of a segment of gold
wire is 100.0 . When the wire is placed in a liquid bath, the resistance
decreases to 97.0 . What is the temperature of the bath?
(Hint: First determine the resistance of the gold wire at room temperature.)

Power Consumed by a
Resistor
E
–
+
V = 0
V = E
I = E/R
R
dU
P
dt
= ( )
d
Q V
dt
= ∆
dQ
V
dt
= ∆ I V= ∆
P I V= ∆
( )
2
2 V
P RI
R
∆
= = ,WATT,W

Solve the current, v-drop across each resistor,
and Req for the circuit.
12v 12v

Example: A rectangular block of iron has dimension 1.2 cm×1.2 cm×15 cm.
(a) What is the resistance of the block measured between the two square ends?
(b) What is the resistance between two opposite rectangular faces?
Solution:
Power:
Example: A wire has a resistance R of 72 Ω. At what rate is energy dissipated
in each of the following situations? (1) A potential difference of 120 V is
applied across the full length of the wire. (2) The wire is is cut in half, and a
potential difference of 120 V is applied across the length of each half.
Solution:
Example: A wire of length L=2.35 m and diameter d=1.63 mm carries a
current i of 1.24 A. The wire dissipates electrical energy at the rate P=48.5
mW. Of what is the wire made?
Solution:

78
Clicker Question
You have three cylindrical copper conductors. Rank them
according to the current through them, the greatest first, when
the same potential difference V is placed across their lengths.
A: a, b, c
B: a and c tie, then b
C: b, a, c
D: a and b tie, then c
B: a and c tie, then b

OHMS LAW
Experimentally, it is found that the current in a wire is
proportional to the potential difference between its
ends:
Ohm’s Law
The ratio of voltage to current is called the
resistance:
or

Microscopic View of Ohm’s Law
The motion of the electrons in an electric field E is a combination of motion due
to random collisions and that due to E. The electron random motions average to
zero and make no contribution to the drift speed. So the drift speed is due only
to the electric field.
In the average mean free time τ between collisions, the average electron will
acquire a speed of
Recalling
that
If an electron of mass m is placed in electron field, it will undergo an acceleration.
, we can write the above result as
Solve for E and we have
Recalling that The above equation becomes

Microscopic Description
of Ohm’s law
•Microscopically current is due to the
movement of charge carriers
When a field is applied, the symmetry of the “motion” of the electrons is
broken and there is a net drift.
τ
m
Enq
J
2
=
nqvJ =
We can rewrite this in a different form,
atv =
m
F
a =qEF =
Put this all together and,
We also now know what the conductivity and
resistivity are.

82
• Coefficient in this
dependence is called
resistance R
• Resistance is measured in
Ohm (Ω = V/A)
Ohm’s law
• Electric current is proportional to voltage.
VI ∝ IRV =
R
V
I

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