This document discusses boundary value problems in electrostatics. It provides the Poisson and Laplace equations, which can be used to solve for electric potential or field when the charge or potential is only known at boundaries. The general procedure is to solve the equations to find the potential, then take derivatives to find the electric field and displacement field. Examples are given to demonstrate applying boundary conditions to determine constants and unique solutions for potential and field.

1. Class 14
Boundary Value Problems

2. Boundary Value Problems
• So far the electric field has been obtained using
Coulomb’s law or Gauss law where the charge
distribution is known throughout the region or by

using−∇V
where the potential distribution
E=
is known. In practical problems the charge or
potential is known only at some boundaries and
it is desired to know the field or potential
throughout the region. Such problems are
tackled using Poisson or Laplace equation.

3. Poisson and Laplace Equations


∇.D = ∇.εE = ρV

E = −∇V
∇.(−ε∇V ) = ρV
ρV
∇ V =−
Poisson Equation
ε
For a ch arg e free region
2
∇ 2V = 0 Laplace Equation

4. Poisson and Laplace Equation
• The Laplace equations in all the 3 coordinate systems
are as given below
∂2 A ∂2 A ∂2 A
in Cartesian coordinates
∇2 A = 2 + 2 + 2
∂x
∂y
∂z
1 ∂
∂A
1 ∂2 A ∂2 A
∇ A=
(ρ ) + 2
+ 2
2
ρ ∂ρ ∂ρ
ρ ∂φ
∂z
2
in cylindrical coordinates
1 ∂ 2 ∂A
1
∂
∂A
1
∂2 A
∇ A= 2 ( r
)+ 2
(sin θ
)+ 2
∂θ
r ∂r ∂r
r sin θ ∂θ
r sin 2 θ ∂φ 2
2
In spherical coordinates

5. General Procedure for solving
Laplace or Poisson Equation:
•
•
•
•
Solve Laplace or Poisson equations for V by (a) direct
substitution for single variable or (b) by method of
separation of variables for more than one variable. The
solution at this point is not unique because of the
integration constants
Apply the boundary conditions to determine the
integration constants giving a unique solution for V.


Having found V, find E = −∇V and D = εE .
If desired find the charge Q induced on a conductor
surface using Q = ∫ ρ s dS and ρ s = Dn where D n is the
component of D normal to the conductor. If necessary
the capacitance between two conductors can be found
using C=Q/V.

6. Practice Example 6.1
• In a one dimensional device, the charge

ρ0 x
density is given by ρ v =
a . If E = 0
at x=0 and V=0 at x=a, find V and E.

7. Solution 6.1
ρv
ρ0 x
∂ 2V
=−
=−
2
∂x
ε
aε
ρ0 x
∂V
=−
+ C1
∂x
2aε
2
ρ0 x3
V =−
+ C1 x + C2
6 aε

8. Solution 6.1
• Substituting V=0 at x=a we get
3
ρ0a
0=−
+ C1a + C2
6 aε

ρ0 x 2
∂V 
E = −∇V = −
ax =
− C1
∂x
2 aε
E = 0 at x = 0;∴ C1 = 0
ρ0a3
ρ0 3 3
(a − x )
C2 =
∴V =
6 aε
6aε
ρ0 x 2 
E=
ax
2aε

9. Practice Exercise 6.3
• Two conducting plates of size 1 x 5m are
inclined at 450 to each other with a gap of
width 4mm separating them as shown in
the figure. Determine an approximate
value of the charge per plate, if the plates
are maintained at a potential difference of
50V. Assume the medium between them
has ε r = 1.5.

10. Solution 6.3
• The potential varies only with respect to .
1 ∂ 2V
∇ 2V = 2
= 0; ρ = 0 is excluded
2
ρ ∂φ
∂ 2V
= 0;V = C1φ + C2
2
∂φ
φ = 0 at V = 0 ⇒ C2 = 0
φ =π
200
at V = 50 ⇒ C1 =
4
π
200
∴V =
φ
π

1 ∂V
200 
E = −∇V = −
=−
aφ
ρ ∂φ
πρ

11. Solution 6.3


300ε 0 
 200  
D = ε 0 ε r E = 1.5 × ε 0 ×  −
 ρπ aφ = − ρπ aφ = ρ s



The gap between the plates is 4mm. Straight lines are
extended from the near ends to meet at O as shown in
the figure. The distance from O to the tip of the
horizontal plate is found as follows.
2
2
sin 22.5 = ; l =
= 5.2265mm
l
sin 22.5
300ε 0
Ch arg e = ∫ ρ s dS = ∫ ∫ ρ s dρdz = −
π
0 0.00523
5
1
dρ
∫ 0.00523 ρ dz = 22.2nC
∫
0
5
1