MIRTA VARGAS DE ARGENTINA MEDIA 9 CALZADA Cat B 2° grupo 1ª Actividad
Tarea_3_Ale_Maestria
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Transcript of Tarea_3_Ale_Maestria
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TEORA DE ELECTRICIDAD Y MAGNETISMOTAREA #3 CHAPTER 2.BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS:I
(28/09/2010)
1
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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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
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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TEORA DE ELECTRICIDAD Y MAGNETISMOTAREA #3 CHAPTER 2.BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS:I
(28/09/2010)
3
Mara Alejandra Llamas Bugarn
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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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
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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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
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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TEORA DE ELECTRICIDAD Y MAGNETISMOTAREA #3 CHAPTER 2.BOUNDARY-VALUE PROBLEMS IN ELECTROSTATICS:I
(28/09/2010)
6
Mara Alejandra Llamas Bugarn
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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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
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: