| Topic: |
Science > Physics |
| User: |
"Sery" |
| Date: |
22 Jul 2006 08:00:31 AM |
| Object: |
apparent paradox with charges |
Can you help me with the following, please?
Why do the work to carry two charges q and -q at a distance " 2a "
maintaining them in symmetrical position with respect to the plane z=0
is the double of the work on each charge while the work necessary in
order to remove at infinity distance a charge Q_1 (the charge Q_2
beeing fixed) is the same as the work needed in order to remove at
infinity distance both Q_1 and Q_2 ?
Thanks, Sery
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| User: "srp" |
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| Title: Re: apparent paradox with charges |
23 Jul 2006 07:39:57 AM |
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Sery a écrit :
Can you help me with the following, please?
Why do the work to carry two charges q and -q at a distance " 2a "
maintaining them in symmetrical position with respect to the plane z=0
is the double of the work on each charge while the work necessary in
order to remove at infinity distance a charge Q_1 (the charge Q_2
beeing fixed) is the same as the work needed in order to remove at
infinity distance both Q_1 and Q_2 ?
Thanks, Sery
Because infinity in this case amounts to an asymptote. Although
mathematically coherent, it is physically impossible to remove
any really existing particle to infinity.
In other words, the math will give an illogical result here simply
because at the limit, it describes something that cannot exist.
André Michaud
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| User: "Greg Neill" |
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| Title: Re: apparent paradox with charges |
23 Jul 2006 08:29:34 AM |
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"srp" <srp2@globetrotter.net> wrote in message news:44C36C71.1020007@globetrotter.net...
Sery a écrit :
Can you help me with the following, please?
Why do the work to carry two charges q and -q at a distance " 2a "
maintaining them in symmetrical position with respect to the plane z=0
is the double of the work on each charge while the work necessary in
order to remove at infinity distance a charge Q_1 (the charge Q_2
beeing fixed) is the same as the work needed in order to remove at
infinity distance both Q_1 and Q_2 ?
Thanks, Sery
Because infinity in this case amounts to an asymptote. Although
mathematically coherent, it is physically impossible to remove
any really existing particle to infinity.
In other words, the math will give an illogical result here simply
because at the limit, it describes something that cannot exist.
If you've managed to puzzle out what it is that the
OP is actually asking, then you might be good enough
to enlighten the rest of us. Otherwise, your answer
is apparently just more gobbledegook.
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| User: "Edward Green" |
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| Title: Re: apparent paradox with charges |
23 Jul 2006 10:12:12 PM |
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Greg Neill wrote:
"srp" <srp2@globetrotter.net> wrote in message news:44C36C71.1020007@glob=
etrotter.net...
Sery a =E9crit :
Can you help me with the following, please?
Why do the work to carry two charges q and -q at a distance " 2a "
maintaining them in symmetrical position with respect to the plane z=
=3D0
is the double of the work on each charge while the work necessary in
order to remove at infinity distance a charge Q_1 (the charge Q_2
beeing fixed) is the same as the work needed in order to remove at
infinity distance both Q_1 and Q_2 ?
Thanks, Sery
Because infinity in this case amounts to an asymptote. Although
mathematically coherent, it is physically impossible to remove
any really existing particle to infinity.
In other words, the math will give an illogical result here simply
because at the limit, it describes something that cannot exist.
If you've managed to puzzle out what it is that the
OP is actually asking, then you might be good enough
to enlighten the rest of us. Otherwise, your answer
is apparently just more gobbledegook.
If you read the OP's statement several times, allowing your eyes to
defocus, you may begin to apprehend... <tinkling sound effect>
"Why, if we remove two charges to infinity keeping them at all times
symmetrical to the plane z =3D 0, is the total work done equal to twice
the work done on either charge, whereas if we keep one charge fixed,
and remove the other to infinity, the work is equal to that done on the
single moving charge, yet equal to the total work in the first case"?
Ah... I don't seem to have done all that much better. Anyway, I claim
priority with the answer to this one, while, on the other hand, SRP's
answer seems to be some, if not more, of the material you mentioned. I
see no relation to the fact that "at infinity" refers to a limiting
value to any possible difficulty here: the "paradox" seems to be that
we reach the same limit if we move one charge off, or both.
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| User: "Greg Neill" |
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| Title: Re: apparent paradox with charges |
24 Jul 2006 06:42:52 AM |
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"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:1153710732.248119.114130@i3g2000cwc.googlegroups.com...
If you read the OP's statement several times, allowing your eyes to
defocus, you may begin to apprehend... <tinkling sound effect>
"Why, if we remove two charges to infinity keeping them at all times
symmetrical to the plane z = 0, is the total work done equal to twice
the work done on either charge, whereas if we keep one charge fixed,
and remove the other to infinity, the work is equal to that done on the
single moving charge, yet equal to the total work in the first case"?
Ah... I don't seem to have done all that much better. Anyway, I claim
priority with the answer to this one, while, on the other hand, SRP's
answer seems to be some, if not more, of the material you mentioned. I
see no relation to the fact that "at infinity" refers to a limiting
value to any possible difficulty here: the "paradox" seems to be that
we reach the same limit if we move one charge off, or both.
Of course, in the case of your "solution" to what the
question is, there can be no doubt that the plane z=0
is entirely arbitrary and serves only as a point of
reference, and one would not expect the choice of
inertial frame to affect the total work calculated...
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| User: "Edward Green" |
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| Title: Re: apparent paradox with charges |
22 Jul 2006 03:49:14 PM |
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Sery wrote:
Can you help me with the following, please?
Why do the work to carry two charges q and -q at a distance " 2a "
maintaining them in symmetrical position with respect to the plane z=0
is the double of the work on each charge while the work necessary in
order to remove at infinity distance a charge Q_1 (the charge Q_2
beeing fixed) is the same as the work needed in order to remove at
infinity distance both Q_1 and Q_2 ?
Because in both cases the charges are in one configuration at distance
2a, in the other at infinite separation, whether or not one remains
near the origin. You could get to the both charges at infinity
configuration from the one charge at infinity configuration by moving
off the remaining charge to infinity without doing additional work, so
if the system is conservative, what you claim must be true.
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| User: "tadchem" |
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| Title: Re: apparent paradox with charges |
22 Jul 2006 11:51:11 AM |
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Sery wrote:
Can you help me with the following, please?
Why do the work to carry two charges q and -q at a distance " 2a "
maintaining them in symmetrical position with respect to the plane z=0
is the double of the work on each charge while the work necessary in
order to remove at infinity distance a charge Q_1 (the charge Q_2
beeing fixed) is the same as the work needed in order to remove at
infinity distance both Q_1 and Q_2 ?
Sorry. I don't understand the question.
Tom Davidson
Richmond, VA
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| User: "Randy Poe" |
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| Title: Re: apparent paradox with charges |
22 Jul 2006 12:52:50 PM |
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tadchem wrote:
Sery wrote:
Can you help me with the following, please?
Why do the work to carry two charges q and -q at a distance " 2a "
maintaining them in symmetrical position with respect to the plane z=0
is the double of the work on each charge while the work necessary in
order to remove at infinity distance a charge Q_1 (the charge Q_2
beeing fixed) is the same as the work needed in order to remove at
infinity distance both Q_1 and Q_2 ?
Sorry. I don't understand the question.
A guess: OP is asking about the work done in putting two
point charges together at distance a, compared to the work
done in assembling those same two charges with an infinite
(conducting?) plane in between.
- Randy
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| User: "Ben Rudiak-Gould" |
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| Title: Re: apparent paradox with charges |
23 Jul 2006 08:58:01 PM |
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Sery wrote:
Why do the work to carry two charges q and -q at a distance " 2a "
maintaining them in symmetrical position with respect to the plane z=0
is the double of the work on each charge while the work necessary in
order to remove at infinity distance a charge Q_1 (the charge Q_2
beeing fixed) is the same as the work needed in order to remove at
infinity distance both Q_1 and Q_2 ?
It sounds like you're asking how the energy needed to remove two charges to
infinity at the same time can be different from the energy needed to remove
them to infinity one by one. The answer is that it isn't.
An exact treatment is complicated because of the retarded potential, so I'll
do the c=infinity case (Newtonian gravity). Say the coordinates of the
particles are f(t) and -g(t), where f(0) = g(0) = a and both functions
increase monotonically to infinity. Then the work to move the particles to
infinity is
int F.d = int{0 to oo} (q^2 / (f(t)+g(t))^2) (f'(t) + g'(t)) dt
which after a change of variable to u = f(t)+g(t) is
int{2a to oo} (q^2 / u^2) du = q^2 / 2a
which is independent of how you move the particles.
-- Ben
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| User: "Edward Green" |
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| Title: Re: apparent paradox with charges |
23 Jul 2006 11:02:28 PM |
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Ben Rudiak-Gould wrote:
Sery wrote:
Why do the work to carry two charges q and -q at a distance " 2a "
maintaining them in symmetrical position with respect to the plane z=0
is the double of the work on each charge while the work necessary in
order to remove at infinity distance a charge Q_1 (the charge Q_2
beeing fixed) is the same as the work needed in order to remove at
infinity distance both Q_1 and Q_2 ?
It sounds like you're asking how the energy needed to remove two charges to
infinity at the same time can be different from the energy needed to remove
them to infinity one by one. The answer is that it isn't.
An exact treatment is complicated because of the retarded potential, so I'll
do the c=infinity case (Newtonian gravity). Say the coordinates of the
particles are f(t) and -g(t), where f(0) = g(0) = a and both functions
increase monotonically to infinity. Then the work to move the particles to
infinity is
int F.d = int{0 to oo} (q^2 / (f(t)+g(t))^2) (f'(t) + g'(t)) dt
which after a change of variable to u = f(t)+g(t) is
int{2a to oo} (q^2 / u^2) du = q^2 / 2a
which is independent of how you move the particles.
I've learned not to question your missives, but your discursion into
"retarded potentials" has me surprised. Who in their right mind would
even dream of doing anything other than a quasi static calculation?
That's not even in the spirit of the thing (electrostatic potential).
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