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Disappearing Field

by lluke9
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lluke9
#1
May4-12, 01:19 PM
P: 27
So say we have a uniform field of magnitude X. Let's say the force that the field exerts is dependent on a particle's mass, M.

Now say we move the particle D distance against the force of the field, XM.
Our potential energy will be equal to XMD, because we exerted XM force over distance D, hence work was converted into potential energy.

Now we suddenly turn off this field, like one would turn off an electric field by discharging a capacitor. What happens to the potential energy?
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Kabbotta
#2
May5-12, 01:50 AM
P: 45
The field you removed carried the energy that creates that potential. So when you remove the source of the field the potential energy will disappear because the energy that caused it was removed from the system. By whatever magic you accomplished that since Gauss wouldn't be happy you removed a source from the system...suddenly ; )

As an example think about the classic problem of what would happen if you could remove the sun from the solar system. Well, after the last gravitational waves from the sun passed the Earth it would no longer have any potential energy from the sun and it would fly off in a straight line at constant velocity having only kinetic energy.

Hope that helps.
tiny-tim
#3
May5-12, 03:46 AM
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hi lluke9!
Quote Quote by lluke9 View Post
Now we suddenly turn off this field, like one would turn off an electric field by discharging a capacitor. What happens to the potential energy?
Quote Quote by Kabbotta View Post
By whatever magic you accomplished that
as Kabbotta indicates, there's no magic way of removing a field!

you discharge a capacitor by connecting the two plates so that electrons flow from one towards the other

that does not happen instantly, it happens very fast, but one electron at a time, and each electron goes through a voltage difference

also, if there's eg a positive charge on a spring between the plates, which was being pulled towards the electrons, each electron also has to overcome the attraction from that charge as it moves away! (so you don't get quite as much energy from discharging the capacitor as you thought you would)

the potential energy "disappearing" between the plates equals the work done by moving the electrons

lluke9
#4
May7-12, 06:41 PM
P: 27
Disappearing Field

Thanks for help-- you've restored my faith in classical mechanics xD

But another question... what if I moved that positive charge away from the capacitor so that it is no longer in the field?
And I moved it in from the top, so it gets potential energy?

Okay, I'm pretty sure I'm doing something wrong, now.
tiny-tim
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May7-12, 06:48 PM
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Quote Quote by lluke9 View Post
... what if I moved that positive charge away from the capacitor so that it is no longer in the field?
you mean, sideways?
then you're moving it through a decreasing electric field
lluke9
#6
May9-12, 03:13 PM
P: 27
But I've done no work, right?
tiny-tim
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May9-12, 03:37 PM
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hmm yes, you're right, there's no work!
i don't know the answer in that case
sophiecentaur
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May9-12, 04:47 PM
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Quote Quote by lluke9 View Post
But I've done no work, right?
This problem is not straightforward but the work done in moving a positive charge from the negative plate to the positive plate can't just disappear! Moving sideways, from the mid point between the plates and out to the side will result in a smaller force acting on the charge but the Potential is defined as the line integral of the force times the distance. Returning to the negative plate will still result in the same amount of work (bigger distance with smaller forces) being got our as was put in initially in moving it the shorter distance (with bigger forces) to the point when it was between the two plates. So the potential remains the same when moving sideways. You have done no work - but 'so what'?


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