Poisson's equation 2nd 4

5,872 views 19 slides Jan 13, 2017
Slide 1
Slide 1 of 19
Slide 1
1
Slide 2
2
Slide 3
3
Slide 4
4
Slide 5
5
Slide 6
6
Slide 7
7
Slide 8
8
Slide 9
9
Slide 10
10
Slide 11
11
Slide 12
12
Slide 13
13
Slide 14
14
Slide 15
15
Slide 16
16
Slide 17
17
Slide 18
18
Slide 19
19

About This Presentation

Poisson's equations


Slide Content

Unit-2
Poission’s equations
JETGI
Mr. Himanshu Diwakar
Assistant Professor
JETGI
1

Poisson’s and Laplace Equations
A useful approach to the calculation of electric potentials
Relates potential to the charge density.
The electric field is related to the charge density by the divergence relationship
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
In a charge-free region of space, this becomes Laplace's equation
JETGI 2

Potential of a Uniform Sphere of Charge
outside
inside
JETGI 3

Poisson’s and Laplace Equations
Poisson’s Equation
From the point form of Gaus's Law
Del_dot_Dr
v
Definition D
DeE
and the gradient relationship
EDelV-
Del_DDel_eE()Del_dot_eDelV( )- rv
Del_DelV
r
v-
e
Laplace’s Equation
if r
v0
Del_dot_D rv
Del_DelLaplacian
The divergence of the
gradient of a scalar function
is called the Laplacian.
JETGI 4

LapR
xx
Vxy,z,( )
d
d
æ
ç
è
ö
÷
ø
d
d yy
Vxy,z,( )
d
d
æ
ç
è
ö
÷
ø
d
d
+
zz
Vxy,z,( )
d
d
æ
ç
è
ö
÷
ø
d
d
+
é
ê
ë
ù
ú
û
: =
LapC
1
rr
r
r
Vrf,z,( )
d
d
×
æ
ç
è
ö
÷
ø
d
d
×
1
r
2ff
Vrf,z,( )
d
d
æ
ç
è
ö
÷
ø
d
d
é
ê
ë
ù
ú
û
×+
zz
Vrf,z,( )
d
d
æ
ç
è
ö
÷
ø
d
d
+:=
LapS
1
r
2r
r
2
r
Vrq,f,( )
d
d
×
æ
ç
è
ö
÷
ø
d
d
×
é
ê
ë
ù
ú
û
1
r
2
sinq()×
q
sinq()
q
Vrq,f,( )
d
d
×
æ
ç
è
ö
÷
ø
d
d
×+
1
r
2
sinq()
2
×
ff
Vrq,f,( )
d
d
d
d
×+:=
Poisson’s and Laplace Equations
JETGI 5

Given
Vxy,z,( )
4y×z×
x
2
1+
:=
x
y
z
æ
ç
ç
è
ö
÷
÷
ø
1
2
3
æ
ç
ç
è
ö
÷
÷
ø
:= eo8.85410
12-
×:=
Vxy,z,( )12=
Find: V @ and rv at P
LapR
xx
Vxy,z,( )
d
d
æ
ç
è
ö
÷
ø
d
d yy
Vxy,z,( )
d
d
æ
ç
è
ö
÷
ø
d
d
+
zz
Vxy,z,( )
d
d
æ
ç
è
ö
÷
ø
d
d
+
é
ê
ë
ù
ú
û
:=
LapR12=
rvLapReo×:= rv1.06210
10-
´=
Examples of the Solution of Laplace’s Equation
D7.1
JETGI 6

Examples of the Solution of Laplace’s Equation
Example 7.1
Assume V is a function only of x – solve Laplace’s equation
V
V
o

d
JETGI 7

Examples of the Solution of Laplace’s Equation
Finding the capacitance of a parallel-plate capacitor
Steps
1 – Given V, use E = - DelV to find E
2 – Use D = eE to find D
3 - Evaluate D at either capacitor plate, D = Ds = Dn an
4 – Recognize that rs = Dn
5 – Find Q by a surface integration over the capacitor plate
C
Q
V
o
eS×
d
JETGI 8

CAPACITANCE
•The ratio of electric charge to electric
potential of a conductor or a device is
called capacitance
•Capacitance C = Q/V
•Unit is farad (F)
•1 farad = 1 coulomb / 1 volt
JETGI 9

PRINCIPLE OF A CAPACITOR
•Capacitor is based on the principle that the
capacitance of an isolated charged conductor
increases when an uncharged earthed conductor is
kept near it and the capacitance is further
increased by keeping a dielectric medium between
the conductors.
JETGI 10

CAPACITANCE OF A PARALLEL PLATE
CAPACITOR
Electric field between the plates,
E = s/e
0
But s=Q/A
\E=Q/Ae
0
Potential difference between the two
plates , V = Ed = Qd/A e
0
Capacitance, C = Q/V
C=A e
0
/d
JETGI 11

CAPACITANCE OF A PARALLEL PLATE CAPACITOR WITH A
DIELECTRIC SLAB
When a dielectric slab is kept between the plates
COMPLETELY filling the gap
E’ = E
0
/K where K is the dielectric constant of the medium.
Potential difference
V’ = E’d = E
0
d/K=Qd/K e
0
A
Capacitance C’ = Q/V’ = K e
0
A/d = KC
\when a dielectric medium is filled between the plates of a
capacitor, its capacitance is increased K times.
JETGI 12

DIELECTRIC STRENGTH
•Dielectric strength of a dielectric is the
maximum electric field that can be applied
to it beyond which it breaks down.
JETGI 13

PRACTICE PROBLEMS
•Calculate the number of electrons in excess
in a body with 1 coulomb of negative
charge.
•Q = ne
•Q = 1C
•e = 1.6 X 10
-19
C
•n = Q/e= 1/(1.6 X 10
-19
C) = 6.25 X 10
18
JETGI 14

Resistance
•The electrical resistance of an electrical conductor is a
measure of the difficulty to pass an electric current through
that conductor.
•The inverse quantity is electrical conductance, and is the
ease with which an electric current passes.
JETGI 15

JETGI 16

RESISTORS IN SERIES AND PARALLEL
nRRRR +++= ...
21eq
RESISTORS IN SERIES
The magnitude of the charge is constant. Therefore, the
flow of charge, current I is also constant.
The potential of the individual resistors are in general
different.
The equivalent resistance of resistors
in series equals the sum of their
individual resistances.

RESISTORS IN PARALLEL
The upper plates of the capacitors are connected together to
form an equipotential surface – they have the same potential.
The lower plate also have equal potential.
The charges on the plates may not necessarily be equal.
RESISTORS IN SERIES AND
PARALLEL
1
21
eq
1
...
11
-
ú
û
ù
ê
ë
é
+++=
n
RRR
R

RESISTORS IN
SERIES AND PARALLEL
JETGI 19
Tags