solutions of electrostatic problems

7/17/2019 Solutions of Electrostatic Problems

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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:

Edmund Li



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:

Edmund Li



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.

Edmund Li



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 .

Edmund Li



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:

Edmund Li



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.

Edmund Li



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:

Edmund Li



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:

Edmund Li



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.

Edmund Li



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

Edmund Li



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:

Edmund Li



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

Edmund Li



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]

Edmund Li



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

Edmund Li



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.

Edmund Li



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

Edmund Li

solutions of electrostatic problems

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