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TEORA DE ELECTRICIDAD Y MAGNETISMOTAREA #3 CHAPTER 2.BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS:I

(28/09/2010)

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Mara Alejandra Llamas Bugarn

Problem 2.1A point charge q is brought to a position a distance d away from an infinite plane

conductor held at zero potential.

Solution:

a) The surface-charge density induced on the plane.

From the symmetry of the problem, we can see that the potential (x) is equivalent to that

produced by the charge q together with an image charge q= -q located a distance don the

opposite side of the plane. Specifically, the potential is given by

In coordinates cartesian the charge is at the point z =(0, 0, d), then the charge density is:

b) The force between the plane and the charge by using Coulomb's law for the force

between the charge and its image;

Directly of Coulombs law:


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c) The total force acting on the plane by integrating

over the whole plane;

Now we have:

The area differential en coordinates spherical, , then:

Changing the variable,

, then:

d) The work necessary to remove the charge q from its position to infinity;

e) The potential energy between the charge q and its image

From the definition of potential energy

This is the energy required to pull the charge q to infinity. The energy is twice that of the

original system because we also have energy associated with the electric field of the image

charge. In the original system, there is only the electric field of a single charge. Furthermore,

notice that the position and hence energy of the image charge depends on the charge and

position ofq.


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Problem 2.2Method of images for to solve the problem of a point charge q inside a hollow,

grounded, conducting sphere of inner radius a.

Solution:

a) The potential inside the sphere;

If the charge is located at the point r, then, by axial symmetry, the image charge must be located

along the direction of r at a distance of r > a. The potential at a point x will be given in the form

For x =a, the potential is zero, so:

Then;

Using the cosines law and of the figure:

b) The induced surface-charge density;

Inside of the sphere:


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c) The magnitud and direction of the force acting on q.

The electric field is:

d)

Is there any change in the solution if the sphere is kept at a fixed potential V? Ifthe sphere has a total charge Q on its inner and outer surfaces?

Keeping the sphere at a fixed non-zero potential requires net charge on the

conducting shell. This can be imaged as a new image at the centre.

Problem 2.3

A straight-line charge with constant linear charge density is locatedperpendicular to the x-y plane in the first quadrant at (x0, y0). The intersecting

planes x = 0, 0 and = 0, x 0 are conducting boundary surfaces held at zero

potential. Consider the potential, fields, and surface charges in the first

quadrant.


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Solution:

a) Verify explicitly that the potential and the tangential electric field vanish on

the boundary surfaces.

The potential can be made to vanish on the specified

boundary surfaces by pretending that we have three image

line charges. Two image charges have charge density and

exist at the locations obtained by reflecting the original

image charge across the x and y axes, respectively. The

third image charge has charge density + and exists at the

location obtained by reflecting the original charge through

the origin.

Adding a plane at y= 0 will then result in additional

image charges for the above two charges: an image of the

original charge giving (x0, -y0) and an image of the initialimage giving+ at(-x0, -y0).

So, the potential:

So:

We may check the validity of this solution by evaluating the potential on the boundary

surface x = 0 and y=0


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We calculate E tangential:

In x=0, we have:

For y=0 the result is similar.

b) Determine the surface charge density on the plane = 0, x 0.

c) Show that the total charge (per unit length in z) on the plane = 0, x 0 is

The total charge is obtained by integrating the surface charge density


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The total charge Qy in the plane x=0 is then:

And the complete charge:

d) Show that far from the origin the leading term in the potential is

Simplify in Taylor expand:

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