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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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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