ELECTROMAGNETICS: Laplace’s and poisson’s equation In Cartesian, cylindrical and spherical coordinates, APPLICATION, examples, Uniqueness theorem

1. ENGIEERING ELECTROMAGNETICS(2151002)
By: Shivangi Singh
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2.  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.
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3. For the Linear material Poisson’s and Laplace’s equation can
be easily derived from Gauss’s equation
𝛻 ∙𝐷 =𝜌𝑉
But,
𝐷 =
∈𝐸
Putting the value of 𝐷 in Gauss Law,
𝛻 ∗
(∈𝐸) =𝜌𝑉
From homogeneous medium for which ∈
is a constant, we write
𝛻 ∙𝐸 = 𝜌𝑉
∈
Also,
𝐸 =−𝛻𝑉
Then the previous equation becomes,
𝜌𝑉
𝛻 ∙(−𝛻𝑉)=
∈
Or,
𝜌𝑉
𝛻 ∙ 𝛻𝑉 =−
∈
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4. 𝛻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. Inthatcase,Poisson’s
equationreduces to, 𝛻2𝑉 =0
 This equation isknownas Laplace’sequation.
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5. 𝜕𝑉 𝑎𝑥 𝜕𝑉 𝑎𝑦 𝜕𝑉 𝑎𝑧
𝛻𝑉 =
𝜕𝑥
+
𝜕𝑦
+
𝜕𝑧
and,
𝛻𝐴 = + +
𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧
𝜕𝑥 𝜕𝑦 𝜕𝑧
Knowing
𝛻2𝑉 =𝛻 ∙𝛻𝑉
Hence, Laplace’s equation is,
2 2 2
𝛻2𝑉 =𝜕 𝑉
+𝜕 𝑉
+𝜕 𝑉
= 0
𝜕𝑥2 𝜕𝑦2 𝜕𝑧2
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6. 𝜕𝑉 1𝜕𝑉 𝜕𝑉
𝛻 ∙𝑉 = 𝑎𝜌 + 𝑎∅+ 𝑎𝑧
𝜕𝜌 𝜌𝜕∅ 𝜕𝑧
and,
𝛻 ∙𝐴 = 𝜌
(𝜌𝐴 ) +
𝜌𝜕𝜌 𝜌 𝜕∅
+
1 𝜕 1𝜕𝐴 𝜕𝐴
∅ 𝑧
𝜕𝑧
Knowing
𝛻2𝑉 =𝛻 ∙𝛻𝑉
Hence, Laplace’s equation is,
𝛻2𝑉 = 1 𝜕
(𝜌𝜕𝑉
) +
𝜌 𝜕𝜌 𝜕𝜌
1 𝜕2𝑉
+𝜕2𝑉
=0
𝜌2 𝜕∅2 𝜕𝑧2
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7. 𝜕𝑉 1𝜕𝑉 1 𝜕𝑉
𝛻 ∙𝑉 =
𝜕𝑟
𝑎𝑟+
𝑟𝜕𝜃
𝑎𝜃 +
𝑟sin 𝜃 𝜕∅
𝑎∅
and,
𝛻 ∙𝐴 =
1
𝑟2𝜕𝑟
2
𝑟
(𝑟 𝐴 ) + 𝜃
𝜕 1 𝜕
(𝐴 sin𝜃) +
1 𝜕𝐴∅
𝑟sin𝜃𝜕𝜃 𝑟sin 𝜃 𝜕∅
Knowing
𝛻2𝑉 =𝛻 ∙𝛻𝑉
Hence, Laplace’s equation is,
𝛻2𝑉 = (𝑟2 ) +
1 𝜕 𝜕𝑉 1 𝜕
𝑟2 𝜕𝑟 𝜕𝑟 𝑟2sin 𝜃𝜕𝜃
(sin𝜃
𝜕𝑉
𝜕𝜃
) +
1 𝜕2𝑉
𝑟2sin 𝜃𝜕∅2
=0
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8. Using Laplace or Poisson’s equation we can obtain:
 Potential at any point in between two surface when potential at two
surface are given.
 We can also obtain capacitance between these two surface.
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9. 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
2 2 2
𝛻2𝑉 =𝜕 𝑉
+𝜕 𝑉
+𝜕 𝑉
= 0
𝜕𝑥2 𝜕𝑦2 𝜕𝑧2
Differentiating given V,
𝜕𝑥
𝜕𝑉
= 2𝑦2𝑧3
𝜕𝑥2
𝜕2𝑉
= 0
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10. 𝜕𝑦 𝜕𝑦2
𝜕2𝑉
= 4𝑥𝑧3
𝜕𝑧
𝜕𝑉
= 6𝑥𝑦2𝑧2
𝜕𝑧2
𝜕2𝑉
= 12𝑥𝑦2𝑧
Adding double differentiating terms,
𝜕2𝑉
+𝜕2𝑉
+𝜕2𝑉
= 0 + 4*𝑧2 + 12*x*𝑦2*z ≠0
𝜕𝑥2 𝜕𝑦2 𝜕𝑧2
Thus given V does not satisfy Laplace equation
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11. 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.
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