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Image Charge Distance from Center of Grounded Sphere
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comunalj@stjohns.edu
science forum beginner


Joined: 15 Jun 2006
Posts: 3

PostPosted: Thu Jun 15, 2006 4:39 pm    Post subject: Image Charge Distance from Center of Grounded Sphere Reply with quote

hi all,

I am an adjunct instructor at St. John's reading my texts again in
hopes of pasing a qualifier so I may continue to persue a PhD in
Quantum Gravity. I started about 2 months ago and realized that I can't
learn physics in a vacuum. Afterall, there is only so far you can go
with only the numerical answer.

As was pointed out to me I should mention that I am not seeking
homework solutions (there are no classes for me - I am learning this on
my own). I need help understanding the concepts behind the problems -
espically the ones I cannot solve.

This question is from Griffiths', Intro. to Electrodynamics, Example
3.2, to wit:

"A point charge q is situated a distance 'a' from the center of a
grounded conducting sphere of radius 'R' (Fig. 3.12). Find the
potential outside the sphere."

Note: a > R. Griffiths' solution does not provide the explanation on
how the value of 'b' (the distance of the image charge, q', which is
inside the sphere b < R) was obtained. The potential I started with is:

V(r) = k(q'/r_1' + q/r_2),

where r'_1 and r_2 are found using the law of cosines from their
respective charges to the observation point outside the sphere and k :=
1/4pi*epsilon_o

The image charge, q', can be found using the fact that the sphere is
grounded:

q' = - (R - a)/( b - R)

I tried using V = 0 at r = infinity, but I get a wrong answer:

b = 2*R - a, (2 times R)

whereas, the correct result is:

b = R^2/a (R-squared).

Thanks in advance for any help you may give me,
-LD
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Igor Khavkine
science forum Guru


Joined: 01 May 2005
Posts: 607

PostPosted: Thu Jun 15, 2006 9:22 pm    Post subject: Re: Image Charge Distance from Center of Grounded Sphere Reply with quote

comunalj@stjohns.edu wrote:

Quote:
This question is from Griffiths', Intro. to Electrodynamics, Example
3.2, to wit:

"A point charge q is situated a distance 'a' from the center of a
grounded conducting sphere of radius 'R' (Fig. 3.12). Find the
potential outside the sphere."

Note: a > R. Griffiths' solution does not provide the explanation on
how the value of 'b' (the distance of the image charge, q', which is
inside the sphere b < R) was obtained. The potential I started with is:

V(r) = k(q'/r_1' + q/r_2),
[...]


First, it's a good idea to get a feel for what kind of problem can be
solved by the method of images. For example, to see when a single image
charge is sufficient, start with the potential written above and try to
see what kind of equipotential surface is described by the equation
V(r) = 0. (Hint: you will find a sphere. Can you express its center and
radius as functions of the magnitude and position of the image charge?)

Once you can answer the above question, it's a simple matter to choose
the position and magnitude of the image charge to match the V(r) = 0
equipotential surface to the location of the grounded sphere given in
the problem.

For reference, this exact problem is treated in Section 2.2 of
Jackson's Classical Electrodynamics.

Hoep this helps.

Igor
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comunalj@stjohns.edu
science forum beginner


Joined: 15 Jun 2006
Posts: 3

PostPosted: Mon Jun 19, 2006 8:47 pm    Post subject: Re: Image Charge Distance from Center of Grounded Sphere Reply with quote

Igor Khavkine wrote:
Quote:
comunalj@stjohns.edu wrote:

This question is from Griffiths', Intro. to Electrodynamics, Example
3.2, to wit:

"A point charge q is situated a distance 'a' from the center of a
grounded conducting sphere of radius 'R' (Fig. 3.12). Find the
potential outside the sphere."


[...]

Quote:
[...]

First, it's a good idea to get a feel for what kind of problem can be
solved by the method of images. For example, to see when a single image
charge is sufficient, start with the potential written above and try to
see what kind of equipotential surface is described by the equation
V(r) = 0. (Hint: you will find a sphere. Can you express its center and
radius as functions of the magnitude and position of the image charge?)

Once you can answer the above question, it's a simple matter to choose
the position and magnitude of the image charge to match the V(r) = 0
equipotential surface to the location of the grounded sphere given in
the problem.

Wonderful. I would <I>never</I> have thought of this. So you not only
gave me a great guide to solving the problem, but you also gave me
a new way to look at image charge problems!

Quote:
For reference, this exact problem is treated in Section 2.2 of
Jackson's Classical Electrodynamics.

Extra doubly super wonderful - the answer!

Quote:
Hoep this helps.

Only slightly ... LOLOL! 8)

thx a <U>billion</U> Igor!

-joe
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ebunn@lfa221051.richmond.
science forum beginner


Joined: 21 Sep 2005
Posts: 38

PostPosted: Mon Jun 19, 2006 8:47 pm    Post subject: Re: Image Charge Distance from Center of Grounded Sphere Reply with quote

In article <1150300021.306100.114320@f6g2000cwb.googlegroups.com>,
<comunalj@stjohns.edu> wrote:

Quote:
As was pointed out to me I should mention that I am not seeking
homework solutions (there are no classes for me - I am learning this on
my own). I need help understanding the concepts behind the problems -
espically the ones I cannot solve.

That was good advice. As a rule, people in this newsgroup love to
help out someone who's working independently and have little patience
with people who seem to be trying to weasel out of doing their homework
for a class.

Quote:
This question is from Griffiths', Intro. to Electrodynamics, Example
3.2, to wit:

"A point charge q is situated a distance 'a' from the center of a
grounded conducting sphere of radius 'R' (Fig. 3.12). Find the
potential outside the sphere."

Note: a > R. Griffiths' solution does not provide the explanation on
how the value of 'b' (the distance of the image charge, q', which is
inside the sphere b < R) was obtained. The potential I started with is:

V(r) = k(q'/r_1' + q/r_2),

where r'_1 and r_2 are found using the law of cosines from their
respective charges to the observation point outside the sphere and k :=
1/4pi*epsilon_o

The image charge, q', can be found using the fact that the sphere is
grounded:

q' = - (R - a)/( b - R)

So far so good!

Quote:
I tried using V = 0 at r = infinity, but I get a wrong answer:

I don't understand what you did here. All values of the unknowns
have V=0 at infinite distance: as you wrote earlier,

Quote:
V(r) = k(q'/r_1' + q/r_2),

and both r_1' and r_2 go to infinity as r goes to infinity. So the
condition V=0 at infinite r doesn't give you an equation for b.

Here's a suggestion. In deriving your equation for q' you made use
of the fact that V=0 at one particular point on the grounded sphere,
namely the point right in between the real charge and the image charge.
Play the same game with a second point on the sphere to get a second equation.

In fact, you should write down the potential at an arbitrary point
on the sphere -- not just a "special" point like the one in between
the two charges. But one way to get the answer is just to consider
two special points where the algebra is easiest. After that you should
verify that your solution really does give V=0 for all points on the sphere.

-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]
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Andrzej Szczepkowicz
science forum beginner


Joined: 19 Jun 2006
Posts: 1

PostPosted: Mon Jun 19, 2006 8:48 pm    Post subject: Re: Image Charge Distance from Center of Grounded Sphere Reply with quote

Igor Khavkine wrote:
Quote:
comunalj@stjohns.edu wrote:


This question is from Griffiths', Intro. to Electrodynamics, Example
3.2, to wit:

"A point charge q is situated a distance 'a' from the center of a
grounded conducting sphere of radius 'R' (Fig. 3.12). Find the
potential outside the sphere."

Note: a > R. Griffiths' solution does not provide the explanation on
how the value of 'b' (the distance of the image charge, q', which is
inside the sphere b < R) was obtained. The potential I started with is:

V(r) = k(q'/r_1' + q/r_2),

[...]

First, it's a good idea to get a feel for what kind of problem can be
solved by the method of images. For example, to see when a single image
charge is sufficient, start with the potential written above and try to
see what kind of equipotential surface is described by the equation
V(r) = 0. (Hint: you will find a sphere. Can you express its center and
radius as functions of the magnitude and position of the image charge?)

Once you can answer the above question, it's a simple matter to choose
the position and magnitude of the image charge to match the V(r) = 0
equipotential surface to the location of the grounded sphere given in
the problem.

For reference, this exact problem is treated in Section 2.2 of
Jackson's Classical Electrodynamics.


The problem is also treated in Feynman's Lectures. Feynman's book is in
general more elementary than Griffiths', while Jackson's is more
advanced than Griffiths'.

If you assume that the equipotential surface is indeed a sphere (if I
recall correctly, Feynman in his Lectures states this without proof),
then you can find both q' and b in the following way: imagine a straight
line going through charge q and the center of the sphere. The line
intersects the sphere in two points, say, P1 and P2. The potential at P1
is 0, and the potential at P2 is 0. This gives two equations for q' and
b, which can be easily solved.

Unfortunately it is a long way from classical electrodynamics to quantum
gravity. I wish I could understand anything from quantum gravity!
Anyway, learning electrodynamics can be useful and fun. Best wishes!
A.S.
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comunalj@stjohns.edu
science forum beginner


Joined: 15 Jun 2006
Posts: 3

PostPosted: Wed Jun 21, 2006 11:42 pm    Post subject: Re: Image Charge Distance from Center of Grounded Sphere Reply with quote

Andrzej Szczepkowicz wrote:
Quote:
Igor Khavkine wrote:
comunalj@stjohns.edu wrote:


This question is from Griffiths', Intro. to Electrodynamics, Example
3.2, to wit:

"A point charge q is situated a distance 'a' from the center of a
grounded conducting sphere of radius 'R' (Fig. 3.12). Find the
potential outside the sphere."

Note: a > R. Griffiths' solution does not provide the explanation on
how the value of 'b' (the distance of the image charge, q', which is
inside the sphere b < R) was obtained. The potential I started with is:

V(r) = k(q'/r_1' + q/r_2),

[...]

First, it's a good idea to get a feel for what kind of problem can be
solved by the method of images. For example, to see when a single image
charge is sufficient, start with the potential written above and try to
see what kind of equipotential surface is described by the equation
V(r) = 0. (Hint: you will find a sphere. Can you express its center and
radius as functions of the magnitude and position of the image charge?)

Once you can answer the above question, it's a simple matter to choose
the position and magnitude of the image charge to match the V(r) = 0
equipotential surface to the location of the grounded sphere given in
the problem.

For reference, this exact problem is treated in Section 2.2 of
Jackson's Classical Electrodynamics.


The problem is also treated in Feynman's Lectures. Feynman's book is in
general more elementary than Griffiths', while Jackson's is more
advanced than Griffiths'.

If you assume that the equipotential surface is indeed a sphere (if I
recall correctly, Feynman in his Lectures states this without proof),
then you can find both q' and b in the following way: imagine a straight
line going through charge q and the center of the sphere. The line
intersects the sphere in two points, say, P1 and P2. The potential at P1
is 0, and the potential at P2 is 0. This gives two equations for q' and
b, which can be easily solved.

Unfortunately it is a long way from classical electrodynamics to quantum
gravity. I wish I could understand anything from quantum gravity!
Anyway, learning electrodynamics can be useful and fun. Best wishes!
A.S.

At another forum the alternate solution (use r = R at theta = 180)
solved the problem quite nicely. So thanks to those who suggested this.
There are three ways to find 'b':
1) V = 0 at 2 special points on the sphere,
2) Compare the two terms in the potential (for the q and q' charges),
set it to zero and compare the terms. (This is Jackson's solution.),
and
3) Igor's solution: work out the surface of the equipotential and solve
for q' and 'b' in terms of all the known parameters ('a' and 'R' and
q). (I haven't done this one yet... I started reading Butkov's
Mathematical Physics.)

Thanks to all ... to quote the Governor of California: "ill b bak..."

-LD
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