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ELECTROSTATIC PROBLEMS Given that all electrostatic mediums are governed by the fact that:
We start off by saying that the electric field is irrotational, and that from the null vector identity, We
express:
We therefore get:
This is Poisson’s equation:
Where is known as the Laplacian operator – the divergence of a gradient of a scalar. The resulting
equation in the different coordinate systems become:
Let us show this for the cylindrical and spherical coordinates.
And similarly:
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In a simple medium where there is no free charge, then Poisson’s equation reduces to Laplace’s equation:
This equation is the governing equation for problems involving a set of conductors maintained at
different potentials. Once V is found, then can be found and then the charge distribution from .An important fact is that solutions to Laplace’s and Poisson’s equations are unique. We shall
neglect to prove this here.
Example
Consider parallel plate capacitors with a separation of d maintained at potentials 0 to . Assuming
negligible fringing determine the potential and surface charge densities on the plates.
We note that a parallel plate capacitor is a situation of a simple medium and thus we use Laplace’s
equation:
Using the Cartesian coordinates then:
Since the y coordinate is the only space variable then:
Now,
Since the electric field opposes the potential rise then:
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METHOD OF IMAGES
The method of replacing bounding surfaces by appropriate image charges in lieu of a formal solution of
Poisson’s or Laplace’s equation is called the method of images.
In order to use this method we must have:
An image charge of opposite polarity
The charges are located in the region of the conducting plane
The conducting plane is equipotential and of infinite extent and depth
We do not want to find the field in the region where image charges are located.
Any given charge configuration above an infinite conducting plane is equivalent tothe combination of given charge configuration and its image configuration with
elimination of the conducting plane.
Example
Consider the point charge shown at a distance d above a grounded xz conducting plane. Find the
voltage at any point P(x,y,z) above the grounded plane.
We first remove the grounded plane and mirror the charge Q to –d with charge –Q. Then:
But note that:
The boundary conditions are:
And the symmetry above the axes:
We can see that our solution satisfies these boundary conditions.
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Example
Consider a point charge Q a distance d above a large grounded conducting plane. Determine the
surface charge density in terms of d and the distance r from the origin which is directly below
the charge Q on the ground plane. Also show that the induced charge on the conducting plane is –
Q.
We remove the ground conducting plane and draw an image charge. At a point r from the origin
directly below Q, we determine that the electric field is only perpendicular:
Now:
Since
Thus:
This is the surface charge, and if we integrate this over the area we get the charge on the plane.
Note that it makes sense to use cylindrical coordinates here and expand from r=0 to infinity and to .
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Example
A positive charge Q is located at distance and from two grounded perpendicular conducting
half planes. Determine the force on Q caused by the charges induced on the planes.
An image charge in the fourth quadrant will make the horizontal half plane potential zero (but not
of the vertical half plane), while the second quadrant image charge will make the vertical half
plane zero (but not of the horizontal plane). The issue is that there is no symmetry, so there must
be an image charge of (+Q) in the third quadrant to satisfy the zero potential boundary condition
on both half planes. Negative charges will be induced on the surfaces.
The net force is the vector summation of each of the coloumb forces:
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LINE CHARGE & PARALLEL CONDUCTORS
Consider a line charge located at a distance d from the axis of a parallel, conducting, circular cylinder of
radius a. They are both infinitely long.
The image line charge must be inside the cylinder to ensure that the cylindrical surface at r=a is
an equipotential surface
Due to symmetry with respect to OP, the image line must like on the line OP.
The method of images requires us to first assume:
We proceed with this assumption and see if the boundary conditions
are met. The electric potential at distance r from a line charge of
density
is determined to be:
Where is some reference point of zero potential. Using this equation, we can determine the potential
at a point on or outside the cylindrical surface by using the line charge and its image:
Note how the reference point disappears in the equation, and as such we don’t care where the locationof is. Although we have already arrived at a solution, we must find values for and r in terms of the
quantities a and d.
Moreover, to ensure that the surface of the cylinder is equipotential,
This requires that the location of be such that we can make similar triangles as shown .
Thus:
Thus:
We can see that the image line charge – can then replace the cylindrical conducting surface, and V and at any point outside the surface can be determined from the line charges of and – . From this, we
can observe that if we have two transmission lines where is not satisfied, we can treat the problem
as an image problem of two infinitely long line charges of opposite polarity.
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The above expression is more accurate since it considers the fact that D may not be significantly greater
than the radius of the conductors. Moreover, Gauss’s law assumes the surface charge distribution is
uniform, which is not strictly correct.
Example
Determine the capacitance per unit length between two long, parallel, circular conducting wires of
radius a. The axes of the wires are separated by distance D.
The equipotential surfaces of the two wires can be considered to have been generated by a pair of
linear changes separated by a distance . The potential difference between the two wires is
given by:
Since a<d, then will end up being positive, and will be negative. Thus:
Now:
Solving the quadratic for d gives us . Thus:
Example
Using the method of images, determine an expression for the capacitance per unit length between
a wire conductor of radius a and an earth plane h metres apart. If the conductivity of the dielectric
is , determine the resistance per unit length between the transmission line and plane.
Removing the earth plane and introducing an image line charge, we find that the electric field
obeys the boundary conditions. By symmetry we use line charges , and we can find the
capacitance of a 2 wire transmission line:
Since we can represent the capacitance of one transmission line to the ground plane as a
capacitor, we can also do so with the other, giving two capacitors in series. Alternatively, the
potential where the ground plane was is half the potential difference between the conductor.
Thus, the capacitance between one conductor with the earth plane is:
Now in the next section we shall see that:
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Example
Determine the capacitance per unit length of a 2 wire transmission line with parallel conducting
cylinders of different radii , their axes being separated a distance of D> .
We assume image charges of and – situated in each cylindrical conductor. The voltage due to
is given by:
Similarly:
Thus:
We must now determine how to express in terms of the radii and distance D. Using similar
triangles we find that:
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POINT CHARGE & CONDUCTING SPHERE
Consider a point charge and a conducting sphere. The problem can thus be represented as
an image charge inside the spherical conductor on the line OQ. However since this
does not create a zero potential surface at R=a. We find that:
Thus:
However,
We call an inverse point of Q with respect to a sphere of radius a.
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BOUNDARY PROBLEMS
Where the method of images cannot be used due to complex geometry, a finite conducting body or only
the potential V in the boundary known, then we must solve Poisson’s or Laplace’s equation. The problems
can be classified as:
Dirichlet: the potential is specified everywhere on the boundary
Neumann: the derivative of potential is specified everywhere on the boundary
Mixed boundary: the potential is specified in some and derivative of potential is specified for the
rest.
The approach to such problems involves first establishing the boundary conditions then:
1. Given V find
2.
Use to find
3.
Evaluate at either capacitor plate
4.
Recognise that
5.
Find Q by surface integration
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CARTESIAN COORDINATES SEPARATION OF VARIABLES
Laplace’s equation in Cartesian coordinates gives us:
We assume that the solution is expressed as:
By substituting this into the equation then, we get:
Divide through by yields:
As an argument, since the differentials only involve their respective variable, then can only contain
and and
. However, since they must add up to be zero, then there is no way in which
any of the differentials can contain any variables – otherwise they wouldn’t cancel! As such we require:
And the condition that must be satisfied is:
The above equations are second order ordinary differential equations, the solutions of which must be one
of the following forms:
Exponential Form
Note that:
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Example
Two grounded semi infinite, parallel plane electrodes are separated by a distance d. A third electrode
perpendicular to them is insulated from both and held at a constant potential . Determine the
potential distribution in the region enclosed by the electrodes.
1.
Establish boundary conditions:
2.
Since:
3.
Determine the function with condition that is complex
4. Determine the function assuming that is positive
5.
Apply boundary conditions:
Also
Thus:
There is one final condition:
There is no way in which the current solution can be constant for variations in y. From linear
algebra, since we know that Laplace’s equation is linear, then the sum of
is also a
solution.
This is a fourier series, where we can evaluate
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Example
The upper and lower conducting plates of a large parallel plate capacitor are separated by d and
maintained at and 0 respectively. Assume that the dielectric constant is with uniform
thickness 0.8d is placed over the lower plate. Determine
a) The potential and electric field intensity in the dielectric slab
b) The potential and electric field in the air space
c) The surface charge densities on the upper and lower plates
Boundary conditions:
This is a one dimensional direchlet problem. Thus:
The general solution is thus:
Since . This is satisfied only if [1] is met:
Thus:
In air similarly:
Now,
Also to satisfy [4]:
To satisfy [3]
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BOUNDARY VALUES PROBLEMS IN CYLINDRICAL COORDINATES
Considering only simple mediums:
In situations where the lengthwise dimension of the cylindrical geometry is large in comparison to its
radius, then we may assume that the field quantities are independent of z such that . The
governing equation of a two dimensional problem then becomes:
By combining the two equations:
Dividing by and multiplying by
To hold for all values or
and
, each term must be a constant and be the negative of the other. Thus:
Example
a)
Then:
b)
In air:
c)
Since
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We note that
Is in the same form as the Cartesian equations. As such, the solutions are the same. We note that if is
unrestricted, k must be an integer, n and the solution is:
Where are constants.
The term, can also be rearranged to give us:
Where n has been written for k, implying that has a range of . The solution is given by:
Thus, we can summarise then:
Depending on the boundary conditions, the complete solution may be a summation of the terms. Note:
When the region of interest includes the cylindrical axis then r=0, then terms containing factor cannot exist
When the region of interest includes the point at infinity, then the terms containing cannot
exist.
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When , then both terms lead to simpler solutions.
And similarly,
The product of the two solutions becomes:
Example
Two infinite plates, insulated from each other are maintained at potentials 0 and as shown in the
figure. Determine the potential distributions for the regions
a)
b)
Since the potentials only change with respect to , this is a one dimensional problem where:
a)
Thus, . At , .
Also,
a)
Similarly, we note that at ,
, So
When k=0
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