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Line integral of Electric field: Electric Potential
Electric field at a field point r, due to a point charge
at origin:

E
1 q
rˆ
2
4 πε0 r
After the discussion of area integral of E over the surface (Flux q/ε0 ), what
about Line integral of E from some point a to some other point b?
b



E  dl  ?
a


 E  dl 
b
a
1
4 πε0
b

a
q
q
dr

r2
4πε0
 1
1 

r  r 

b 
 a
b
 
   E  dl  V b   V a 
a
And if a=∞, b=r
r 


V r     E  dl

The electric potential at a distance r from the point charge is the work done per unit
charge in bringing a test charge from infinity to that point.
The integral around a closed path is zero
 
  E  dl  0
1
Relation between E and V


E  V
V (r )
or 
r b 

 V b   V a    V  dl
As fundamental theorem for gradient is
a
 Surface over which Potential is constant is called an equi-potential surface.
Reference point :
convention at infinity.
V V=V
 +V
 +…..
0
Superposition principle:
1
2
Unit: Nm/C or Joule/C or VOLT
Curl of E ?
 
  E  dl  0 The integral around a closed path is zero
 
 

Using Stokes’ theorem =>     E  da   E.dl  0

Hence curl of E

 s 
 E  0
l
In electrostatics only means
no moving charge or current2
Electric Potential due to a Point Charge
Electric Field:
Q
E  ke 2 (radially outward)
r
Electric Potential:

ds
 
V  Vb  Va    E  ds
rb
ra
rb
rb
dr
   Edr  keQ  2
r
ra
ra
1 1
 keQ   
 rb ra 
Convention: V=0 at infinite r
Q
V  ke (electric potential)
r
3
Electric Potential due to a Point Charge
Electric potential at a
distance r from a
positive charge Q
Electric potential at a
distance r from a
negative charge Q
4
Electric Potential due to a System of
Point Charges
For a system of point charges Qi at distances ri
from a point P:
Q1
r1
Qi
V  ke 
ri
i
P
r2
Q2
r4
Q4
r3
Q3
… an algebraic sum of scalars!
5
Electric Potential due to a
Continuous Charge Distribution
If the charges are not discrete but are
continuously distributed over some object or
region, the summation is replaced by an integral:
Qi
V  k
i ri
dq
 V  k
r
where r is the distance from a point P where the
potential is to be determined to an element of
charge dq and the integral is taken over the entire
distribution of charge.
6
Example of Continuous Charge Distribution:
Ring of Charge
Calculate the electric potential V at a distance x
along the axis of a thin, uniformly charged ring of
radius R carrying a total charge Q.
dq
r
V  ke 
dq
 ke 


x2  R2
ke
x R
keQ
2
2
x2  R2
 dq
Uniformly Charged Ring
7
Example of Continuous Charge Distribution:
Uniformly Charged Disk
Calculate the potential on the axis at a distance x from a uniformly
charged disk of radius R.
Divide the disk into thin rings of radius r and thickness dr.
dq
2 rdr

Q
 R2
 2Q 
dq   2  rdr
R 
V  ke 
2
2

r  x r
dq
V  ke 
r
R
dq
x2  r 2
2k e Q
V 
2
R
2Q
 ke 2 
R 0
x r
2
2
rR
r 0
rdr
x2  r 2
2k e Q

2
R

x R x
2
2
8

Determination of the Electric Field from
the Electric Potential
If the electric potential is known in space, the
electric field may be determined from it.
Relationship between electric potential and electric field:
 
VBA  VB  VA    E  ds
B
A
dV
Es  
ds
The component of E in any direction is the negative of the
rate of change of theelectric potential with the distance in
that direction.of E in some arbitrary direction
(component
9

s)
Potential for a Spherical Shell
of Charge Q
keQ
E 2
r
R
r
r
 
dr
V    E  ds  keQ  2
r

r
Q
 1
Vr  V  keQ    ke
r
 r 
On surface of sphere (r = R),
Q
V  ke
R
V must be continuous, so inside
the shell (r < R),
Q
V  ke
R
10
Potential for a Spherical Shell
of Charge
Outside:
Inside:
Inside the shell, there is no electric field, so
it takes no work to move the test charge
around inside the shell, therefore the
electric potential is the same everywhere
inside the shell. V=kQ/R
11
Electric Potential and
Electric Field of a Hollow
Conducting Sphere of
Radius R and Net Charge
Q
How would the electric
potential and electric
field compare for a solid
conducting sphere of the
same radius with the
same net charge?
12
Find the potential inside and outside a uniformly charged solid sphere of radius
R and total charge q. Use infinity as your reference point. Sketch V(r) .
E 
Electric field at r > R:
kQ
r2
Electric potential at r > R:
r
V  

kQ
r2
dr 
kQ
r
E 
Electric field at r < R:
Electric
potential
at r < R:
R
V  

kQ
r
2
R
kQr
R3
r
dr  
R

kQ
R
3
rdr

kQ
kQ 2
kQ 
r2
2
V 

r R 
3 2
3

R
2R 
2R
R
PGGC -11 DR Bhandari




13
Equipotential Surfaces and Lines
• The electric potential is a scalar characteristic of
the electric field (a vector quantity).
• Regions of space at the same electric potential
are called equipotential surfaces.
Uniform Electric Field
Point Charge
Equipotential (blue dashed) lines indicate where the
equipotential surfaces intersect the page and are always
perpendicular to the electric field E.
14
Equipotential surfaces are always
perpendicular to electric field lines.
15
Electric Potential Energy
The change ΔU in the potential energy is given by
ΔU = - Wc = - ∫Fcds
Relating this to a force exerted by an electric field E on a point charge qo is
Fe = qo E
We define the change ΔU in the electric potential energy as
ΔU = - ∫ qoE ds
16
- Wc = ΔU = - ∫ qoE ds
The electric potential energy depends
upon the charge placed in the electric
field. To quantify the potential energy
in terms of only the field itself it is
more useful to define it per unit charge
-Wc/qo =
 
U
V 
   E  ds
q0
A
B
17
Poisson's and Laplace's Equation
• The electric field is related to the charge density by the
divergence relationship
and the electric field is related to the electric potential by a gradient
relationship
• Therefore the potential is related to the charge density by Poisson's
equation
where DEL SQUARE is called Laplacian operator.
• In a charge-free region of space, this becomes Laplace's
equation
18
Summary:
Conversion from one to another
ρ

E
1   dv
aˆ r
2
4 0 r
E
.E 

0
 2V  
E  V

0

1   dv
V
aˆ r
4 0 r
V
r
  E.dl  V

19
Potential Difference (voltage)
r
• in terms of E:
Edl  0
V (r)  V (a)   E  d l
and
a
• Spherical symmetry: V = V (r)
dV (r )
E(r )  
rˆ
dr
where r  r
• Potential energy: U = q V (joules)
• Equi-potential surfaces: perpendicular to field
lines
20
Electric Dipole
• finite dipole
q 1 1
  
V (r ) 
40  r r 
1
1
s

 1  cos q 
as r  
r
r  2r

1 1
s
and
  2 cos q
r r
r
qs 1
+q
V (r ) 
cos q
2
40 r
r+
r
s
q
r-q
p = qs is the dipole moment for a finite dipole
21
Multipole expansion
• Legendre polynomials: coefficients of 1/ r
expansion in powers of   r ' / r  1
r  r  r'  r  r ' 2 r  r'  r 1  
2
2
where    2  2 cos q
and
1
3 2
1 / 2
(1   )
 1      ...
2
8
2

3
cos
q 1 
2
  ...
 1   cos q   
2


22
• Multipole expansion of V:
using
1
1  r' 
1 rˆ  r '
    Pl (cos q ' )   2  ...
r l 0 r  r 
r r
1
1
l
V (r ) 
r ' Pl (cos q ' ) dq'

l 1 
40 l 0 r
l
1  Q p  rˆ

V (r) 
  2  ...
40  r
r

Q   dq'
and
with
p   r ' dq'
23
Linear Dielectrics
• D, E, and P are proportional to each other for
linear materials
0

k
ce


 / 0
k1
D   E  k  0E
permittivity space
permittivity material
dielectric constant
susceptibility
C  k C0
P  D   0E  (   0 )E   0 c e E
24
Capacitors (vacuum)
• Capacitor: charge +Q and –Q on each plate
- - - - - d
E
++++++

Q
d
E 
, so V  E d  Q
 0 A 0
A 0
• define capacitance C :
with units 1F = 1C /1V
Q = C V
C
0 A
d
25
Energy in dielectrics
• vacuum energy density: u (r )  1 ( E)  E
0
2
• dielectric: u (r )  1 D  E
2
• Energy stored in dielectric system:
1
W   D  E d
2
• energy stored in capacitor:
1
k
2
2
W  CV  C0V
2
2
26
• charging a capacitor C: move dq at a given
time from one plate to the other adding to the
q already accumulated
2
2
q
1Q
CV
dW  dq 
W

C
2 C
2
• in terms of the field E
1 A
1
2
2
W   0 ( Ed )   0 E (Vol)
2 d
2

u (r ) 
0
2
2
E (r )
energy density
27
Capacitors
1. Capacitors are devices that store electric charge
2. The capacitor is the first example of a circuit element
A circuit generally consists of a number of electrical components
(called circuit elements) connected together by conducting wires
forming one or more closed loops
Definition of Capacitance
The capacitance, C, of a capacitor is defined as the ratio of the magnitude of the charge
on either conductor to the potential difference between the conductors

Q
C
V
C is always positive
 The SI unit of capacitance is a farad (F)
28
Makeup of a Capacitor
-Q
 A capacitor is basically two parallel conducting
plates with air or insulating material in between.
 A potential difference exists between the conductors
due to the charge
 The capacitor stores charge


E


V0
+Q
d



V
1
29
More About Capacitance
 Capacitance will always be a positive quantity
 The capacitance of a given capacitor is constant
 The capacitance is a measure of the capacitor’s ability to store charge
 The Farad is a large unit, typically you will see microfarads (mF) and
picofarads (pF)
 The capacitance of a device depends on the geometric arrangement of
the conductors
Metallized Polyester Film
Capacitors
Ceramic Capacitors
Polystyrene
Film Capacitors
Electrolytic Capacitors
Electric Double
Layer Capacitors
(Super Capacitors)
Variable Capacitors
30
Capacitance – Isolated Sphere
Q
Q
R
C


 4 o R
V keQ / R ke
Capacitance – Parallel Plates
o A
Q
Q
Q
C



V Ed Qd /  o A
d
Energy Stored in a Capacitor
Q2 1
1
U
 QV  C( V )2
2C 2
2
31
Capacitors in Series
Vab
C1
a
+Q -Q
V1
A
C2
B
C3
Ceq
b
+Q -Q
V3
+Q -Q
V2
+Q -Q
V
+ V
1 Q  C1V1 Q  C2 V2
+ V
Q  C3 V3
2 Vab  V  V1  V2  V3
Substituting for V1, V2, and V3: V 
3 Q  Ceq V
Q Q Q


C1 C2 C3
Substituting for V:
Q
Q Q Q



Ceq C1 C2 C3
Dividing both sides by Q:
1
1
1
1



Ceq C1 C2 C3
Equivalent capacitance
1
1

Ceq
i Ci
32
C1
Q1
a
Capacitors in Parallel
Ceq
a
Q
C2 -
+
Q2
C3
+ -
Q3
+ -
V
V
1 Q1  C1V Q2  C2 V Q3  C3 V
2 Vab  V  V1  V2  V3
3 Q1  Q2  Q3  Ceq V
Substituting for Q1, Q2, and Q3: CeqV  C1V  C2V  C3V
Dividing both sides by V:
Equivalent capacitance
Ceq  C1  C2  C3
Ceq   Ci
i
33