Unit-2
Poission’s equations
JETGI
Mr. Himanshu Diwakar
Assistant Professor
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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
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Potential of a Uniform Sphere of Charge
outside
inside
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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.
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LapR
xx
Vxy,z,( )
d
d
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ç
è
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d
d yy
Vxy,z,( )
d
d
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d
d
+
zz
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d
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d
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é
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: =
LapC
1
rr
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d
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1
r
2ff
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×+
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LapS
1
r
2r
r
2
r
Vrq,f,( )
d
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d
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1
r
2
sinq()×
q
sinq()
q
Vrq,f,( )
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d
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sinq()
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ff
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×+:=
Poisson’s and Laplace Equations
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Given
Vxy,z,( )
4y×z×
x
2
1+
:=
x
y
z
æ
ç
ç
è
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÷
÷
ø
1
2
3
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:= eo8.85410
12-
×:=
Vxy,z,( )12=
Find: V @ and rv at P
LapR
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:=
LapR12=
rvLapReo×:= rv1.06210
10-
´=
Examples of the Solution of Laplace’s Equation
D7.1
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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
x×
d
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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
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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
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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.
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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
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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.
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DIELECTRIC STRENGTH
•Dielectric strength of a dielectric is the
maximum electric field that can be applied
to it beyond which it breaks down.
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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
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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.
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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
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é
+++=
n
RRR
R