poisson’s and laplace’s equation
TRANSCRIPT
Gandhinagar Institute
Of Technology
Subject – Field Theory (2140909)
Branch – Electrical
Topic – Poisson’s and Laplace’s Equation
Name Enrollment No.
Abhishek Chokshi 140120109005
Himal Desai 140120109008
Dheeraj Yadav 140120109010
Guided By – Prof. Supraja Ma’am
Laplace’s and Poisson’s Equation
We have determined the electric field 𝐸 in a region using
Coulomb’s law or Gauss law when the charge distribution is
specified in the region or using the relation 𝐸 = −𝛻𝑉 when the
potential V is specified throughout the region.
However, in practical cases, neither the charge distribution nor
the potential distribution is specified only at some boundaries.
These type of problems are known as electrostatic boundary
value problems.
For these type of problems, the field and the potential V are
determined by using Poisson’s equation or Laplace’s equation.
Laplace’s equation is the special case of Poisson’s equation.
For the Linear material Poisson’s and Laplace’s equation can
be easily derived from Gauss’s equation
𝛻 ∙ 𝐷 = 𝜌𝑉But,
𝐷 =∈ 𝐸
Putting the value of 𝐷 in Gauss Law,
𝛻 ∗ (∈ 𝐸) = 𝜌𝑉From homogeneous medium for which ∈ is a constant, we write
𝛻 ∙ 𝐸 =𝜌𝑉∈
Also, 𝐸 = −𝛻𝑉
Then the previous equation becomes,
𝛻 ∙ (−𝛻𝑉) =𝜌𝑉∈
Or,
𝛻 ∙ 𝛻𝑉 = −𝜌𝑉∈
𝛻2𝑉 = −𝜌𝑉∈
This equation is known as Poisson’s equation which state that
the potential distribution in a region depend on the local
charge distribution.
In many boundary value problems, the charge distribution is
involved on the surface of the conductor for which the free volume charge density is zero, i.e., ƍ=0. In that case, Poisson’s equation reduces to,
𝛻2𝑉 = 0
This equation is known as Laplace’s equation.
Laplace Equation in Three Coordinate
System
In Cartesian coordinates:
𝛻𝑉 =𝜕𝑉
𝜕𝑥𝑎𝑥 +
𝜕𝑉
𝜕𝑦𝑎𝑦 +
𝜕𝑉
𝜕𝑧𝑎𝑧
and,
𝛻𝐴 =𝜕𝐴𝑥𝜕𝑥
+𝜕𝐴𝑦
𝜕𝑦+𝜕𝐴𝑧𝜕𝑧
Knowing
𝛻2𝑉 = 𝛻 ∙ 𝛻𝑉
Hence, Laplace’s equation is,
𝛻2𝑉 =𝜕2𝑉
𝜕𝑥2+𝜕2𝑉
𝜕𝑦2+𝜕2𝑉
𝜕𝑧2= 0
In cylindrical coordinates:
𝛻 ∙ 𝑉 =𝜕𝑉
𝜕𝜌𝑎𝜌 +
1
𝜌
𝜕𝑉
𝜕∅𝑎∅ +
𝜕𝑉
𝜕𝑧𝑎𝑧
and,
𝛻 ∙ 𝐴 =1
𝜌
𝜕
𝜕𝜌(𝜌𝐴𝜌) +
1
𝜌
𝜕𝐴∅𝜕∅
+𝜕𝐴𝑧𝜕𝑧
Knowing
𝛻2𝑉 = 𝛻 ∙ 𝛻𝑉
Hence, Laplace’s equation is,
𝛻2𝑉 =1
𝜌
𝜕
𝜕𝜌(𝜌𝜕𝑉
𝜕𝜌) +
1
𝜌2𝜕2𝑉
𝜕∅2+𝜕2𝑉
𝜕𝑧2= 0
In spherical coordinates:
𝛻 ∙ 𝑉 =𝜕𝑉
𝜕𝑟𝑎𝑟 +
1
𝑟
𝜕𝑉
𝜕𝜃𝑎𝜃 +
1
𝑟 sin 𝜃
𝜕𝑉
𝜕∅𝑎∅
and,
𝛻 ∙ 𝐴 =1
𝑟2𝜕
𝜕𝑟(𝑟2𝐴𝑟) +
1
𝑟 sin 𝜃
𝜕
𝜕𝜃(𝐴𝜃 sin 𝜃) +
1
𝑟 sin 𝜃
𝜕𝐴∅𝜕∅
Knowing
𝛻2𝑉 = 𝛻 ∙ 𝛻𝑉
Hence, Laplace’s equation is,
𝛻2𝑉 =1
𝑟2𝜕
𝜕𝑟(𝑟2𝜕𝑉
𝜕𝑟) +
1
𝑟2 sin 𝜃
𝜕
𝜕𝜃(sin 𝜃
𝜕𝑉
𝜕𝜃) +
1
𝑟2 sin 𝜃
𝜕2𝑉
𝜕∅2= 0
Application of Laplace’s and Poisson’s
Equation
Using Laplace or Poisson’s equation we can obtain:
1. Potential at any point in between two surface when
potential at two surface are given.
2. We can also obtain capacitance between these two
surface.
EXAMPLE:
Let 𝑉 = 2𝑥𝑦3𝑧3and ∈=∈0. Given point P(1,3,-1).Find V at
point P. Also Find V satisfies Laplace equation.
SOLUTION:
𝑉 = 2𝑥𝑦3𝑧3
V(1,3,-1) = 2*1*32(−1)3
= -18 volt
Laplace equation in Cartesian system is
𝛻2𝑉 =𝜕2𝑉
𝜕𝑥2+𝜕2𝑉
𝜕𝑦2+𝜕2𝑉
𝜕𝑧2= 0
Differentiating given V,
𝜕𝑉
𝜕𝑥= 2𝑦2𝑧3
𝜕2𝑉
𝜕𝑥2= 0
𝜕𝑉
𝜕𝑦= 4𝑥𝑦𝑧3
𝜕2𝑉
𝜕𝑦2= 4𝑥𝑧3
𝜕𝑉
𝜕𝑧= 6𝑥𝑦2𝑧2
𝜕2𝑉
𝜕𝑧2= 12𝑥𝑦2𝑧
Adding double differentiating terms,
𝜕2𝑉
𝜕𝑥2+𝜕2𝑉
𝜕𝑦2+𝜕2𝑉
𝜕𝑧2= 0 + 4*𝑧2 + 12*x*𝑦2*z ≠ 0
Thus given V does not satisfy Laplace equation
EXAMPLE:
Consider two concentric spheres of radii a and b, a<b. The
outer sphere is kept at a potential 𝑉0 and the inner sphere at
zero potential. Solve Laplace equation in spherical
coordinates to find,
1. The potential and electric field in the region between two
spheres.
2. Find the capacitance between them.
SOLUTION:
From the given data it is clear that V is a function of r only.
i.e. 𝜕𝑉
𝜕𝜃and
𝜕𝑉
𝜕𝜙are zero. We know the Laplace's equation in
spherical coordinate system as,
𝛻2𝑉 =1
r2𝜕
𝜕𝑟𝑟2𝜕𝑉
𝜕𝑟+
1
𝑟2 sin 𝜃
𝜕
𝜕𝜃sin 𝜃
𝜕𝑉
𝜕𝜃+
1
𝑟2 sin 𝜃
𝜕2𝑉
𝜕𝜙2= 0
This equation reduces to,
𝛻2𝑉 =1
r2𝜕
𝜕𝑟𝑟2𝜕𝑉
𝜕𝑟= 0
𝑖. 𝑒.𝜕
𝜕𝑟𝑟2𝜕𝑉
𝜕𝑟= 0
By Integrating we get,
𝑟2𝜕𝑉
𝜕𝑟= 𝐴
Or,𝜕𝑉
𝜕𝑟=𝐴
𝑟2
Integrating again we get,
𝑉 = −𝐴
𝑟+ 𝐵…………………(i)
The given boundary condition are,
V = V0 at r = b and V = 0 at r = a
Applying these boundary con to equ (i)
V0 = −A
b+ B and 0 = −
A
a+ B…………(ii)
Subtracting equ (i) from equ (ii)
V0 − 0 = −A
b+A
a= A
1
a−1
b
A =+V01a−1b
Putting value of A in equ (ii) we get,
B =+V0b
1a−1b
Putting these values in equ (i), we get potential V as,
𝑉 =𝑉0
𝑟1𝑎−1𝑏
+𝑉0
𝑏1𝑎−1𝑏
= 𝑉0
1𝑟+1𝑏
1𝑎−1𝑏
We have,
𝐸 = − 𝛻 𝑉 = −𝜕𝑉
𝜕𝜌𝑎𝑟
𝐸 =𝑉0
𝑟21𝑎−1𝑏
𝑎𝑟 ( 𝑉/𝑚)
(b) To Find capacitance :
The flux density is,
𝐷 = ∈ 𝐸 =∈ 𝑉0
𝑟21𝑎−1𝑏
𝑎𝑟 (𝐶/𝑚2)
The charge density on the inner sphere i.e. at r=a is,
𝜌𝑠 = 𝐷 | 𝑟 = 𝑎 ∙ 𝑎𝑟 =∈ 𝑉0
𝑎21𝑎−1𝑏
(𝐶/𝑚2)
The capacitance is now calculated by,
𝐶 =𝑄
𝑉0=
4 𝜋 𝑎2𝜌𝑠
𝑉0=
4𝜋∈1
𝑎−1
𝑏
(𝐹)…...Ans
EXAMPLES:
Two parallel conducting disks are separated by a distance
5mm at Z=0 and Z=5 mm. If V=0 at Z=0 and V=100 at
z=5 mm. Find charge densities on the disks.
SOLUTION:
From the given con. It is clear that V is a fun. Of z only.
Laplace’s equation of cylindrical coordinate system, we get
𝛻2𝑉 =𝜕2V
𝜕z2= 0; 𝑖. 𝑒
𝜕2V
𝜕z2= 0
Integrating twice we get,dV
d2= A and V = Az + B …… i
Where, A and B are constants of integration. To determine
these constants apply the boundary conditions.
V=0 at z=0 and V=100 Volts at z=5 mm
Putting these in equ (i)
0 = 𝐴 0 + 𝐵 𝑎𝑛𝑑 100 = 𝐴 5 × 10−3 + 𝐵
𝐵 = 0 𝑖. 𝑒. 𝐴 =100
5 × 10−3= 20 × 103
Putting these in equ (i), the potential is obtained as,
𝑉 = 20 × 103𝑧
Now, the electric field between the circular disks is obtained
using the relation,
𝐸 = − 𝛻 𝑉 = −𝜕𝑉
𝜕𝜌𝑎𝜌 +
𝜕𝑉
𝜌 𝜕∅𝑎∅ +
𝜕𝑉
𝜕𝑧𝑎𝑧
E = −𝜕V
𝜕z az = − 20 × 103 az
consider, the medium between the circular disks as free
space, which gives the flux density as,
D = ∈0 E =10−9
36 π× −20 × 103 az
= −0.177 × 10−6 az (C /m2 )
Since D is constant between the phase and normal to the
plates, we say that
ρs = Dn = ±0.177 × 10−6 az (C /m2 )
Uniqueness Theorem
STATEMENT:
A solution of Poisson’s equation (of which
Laplace’s equation is a special case) that satisfies the given
boundary condition is a unique solution.
PROOF:
Let us assume that we have two solution of Laplace’s
equation, 𝑉1 and 𝑉2, both general function of the coordinate
use. Therefore,
𝛻2𝑉1 = 0
and
𝛻2𝑉2 = 0
Also assume that both 𝑉1 and 𝑉2 satisfies the same
boundary condition . Let us define a new difference
potential :
𝑉1 − 𝑉2 = 𝑉𝑑Now
𝛻2(𝑉1 − 𝑉2) = 0
i.e
𝛻2𝑉𝑑 = 0
We have the vector identity,
𝛻 ∙ 𝑉𝐷 = 𝑉 𝛻 ∙ 𝐷 + 𝐷 ∙ (𝛻𝑉)
Let V= 𝑉𝑑 and D= 𝛻 𝑉𝑑 , then we have
𝛻 ∙ 𝑉𝑑𝛻 𝑉𝑑 = 𝑉𝑑 𝛻2𝑉𝑑 + 𝛻𝑉𝑑
2
From the above equation, 𝛻2𝑉𝑑 = 0, then we have
𝛻 ∙ 𝑉𝑑𝛻 𝑉𝑑 = 𝛻𝑉𝑑2
Taking the volume integral over the volume gives,
𝑉
𝛻 ∙ 𝑉𝑑𝛻 𝑉𝑑 𝑑𝑣 = 𝑣
𝛻𝑉𝑑2 𝑑𝑣
Now the divergent theorem is:
𝑠
𝐷𝑑 𝑠 = 𝑣
(𝛻 ∙ 𝐷)𝑑𝑣
The right hand side of above equation can now be written in
term of surface integral as,
𝑠
𝑉𝑑𝛻 𝑉𝑑 ∙ 𝑑 𝑠 = 𝑣
𝛻𝑉𝑑2 𝑑𝑣
The surface integral can be evaluated by considering the
surface of very large sphere with radius R. when R is very
large both 𝑉1 and 𝑉2 can be treated as a point charge,
since size of it will be very small compared to R. Hence
the surface integral on LHS decreases and approaches
zero. Thus we have,
𝑣
𝛻𝑉𝑑2 𝑑𝑣 = 0
Since 𝛻𝑉𝑑2 is always positive, everywhere above
equation is satisfied only if 𝛻𝑉𝑑=0. If the gradient of 𝑉𝑑 i.e.
𝑉1 − 𝑉2 is everywhere zero, then 𝑉1 − 𝑉2 cannot change
with any coordinates i.e. it has same value at all point on
its boundary surface.
Therefore, 𝑉1 = 𝑉2 gives two identical solutions, i.e. unique
solution
