4'1 The method of images

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4.1 The method of images We have found that we
can find a solution to Poissons Equation for a
charge q above a conducting (V0) plane by
removing the plane and replacing it with a charge
of q an equal distance below the location of the
plane
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Using this concept we can also find the force
between the charge and the (charge induced in
the) conducting plane
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Using this concept we can also find the force
between the charge and the (charge induced in
the) conducting plane However, when we try
to use this idea to calculate the electrostatic
energy of this configuration we need to be
careful. At first we might try to directly
calculate this by finding the energy of the
charge and its image charge
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Using this concept we can also find the force
between the charge and the (charge induced in
the) conducting plane However, when we try to
use this idea to calculate the electrostatic
energy of this configuration we need to be
careful. At first we might try to directly
calculate this by finding the energy of the
charge and its image charge We get a result
that is a factor of two too large. If we
calculate the energy by moving the real charge in
from infinity to d away from the plane we get the
correct result
5
Using this concept we can also find the force
between the charge and the (charge induced in
the) conducting plane However, when we try to
use this idea to calculate the electrostatic
energy of this configuration we need to be
careful. At first we might try to directly
calculate this by finding the energy of the
charge and its image charge We get a result
that is a factor of two too large. If we
calculate the energy by moving the real charge in
from infinity to d away from the plane we get the
correct result
6
This technique can be applied in other
situations. Consider the following, even more
awkward, situation involving a grounded
conducting sphere
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This technique can be applied in other
situations. Consider the following, even more
awkward, situation involving a grounded
conducting sphere By placing a charge
q a distance a away from the centre of the
conducting sphere, we can construct a situation
that meets the boundary conditions of the
original problem
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4.2 Separation of Variables (Griffiths 3.3) We
will now take a more general approach to finding
a solution to Laplaces Equation - the method of
separation of variables. The general principle
is that we will look for solutions that are
products of functions, each of which depends on
only one of the coordinates. In general although
this is a fairly simple idea it is mathematically
quite messy.
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Consider the following problem With the
boundary conditions V 0 when y 0 V 0 when
y a V V0(y) when x 0 V ? 0 as x ? 8 We
require a solution to i.e. the solution is
independent of z.
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What we will look for is a solution of the form
of a product of functions This, at face
value, seems like a crazy restriction to place on
our solution as very few solutions to Laplaces
Equation have this form! This implies that
If we then divide through by our expression for
V we can then separate the variables
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So now we have two terms, each dependant on only
one variable in other words it is of the
form Now, for a general x and y the only way
that this can be true is if each of these terms
is equal to a constant they can then cancel and
sum to zero So, if one of the constants is
positive, the other must be negative and vise
versa. For each particular problem you need to
check which combination is appropriate. We will
see in a moment that for this problem C1 is
positive and C2 is negative and therefore
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Notice that we have now reduced our partial
differential equation to two ordinary
differential equations, and ones we know how to
solve too!
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Notice that we have now reduced our partial
differential equation to two ordinary
differential equations, and ones we know how to
solve too! so Now we have constructed this as
a solution to Laplaces Equation so all we need
to do now is see if we can determine what
constants we need to use to meet the boundary
conditions.

• V 0 when y 0
• V 0 when y a
• V V0(y) when x 0
• V ? 0 as x ? 8