"compression" of an electrostatic charged sphereby cala Tags: charged, compression, electrostatic, sphere 

#1
May303, 05:17 AM

P: 176

Everytime i propose a new motion machine from my ignorance or misunderstunding of physics, your explanations solve some questions, but open new others:
I proposed a system that acumulated electrons on a sphere surrounded by an electrons charged outter sphere, then connecting them, all the electrons will pass to the outter sphere. Now I know that the outter sphere will create a voltage into the cavity that will not let the electrons go into the inner sphere, unless you force them aplying energy. (Urkil knows how hard it was to explain me that). OK, so the electrons on the outter sphere will create a constant voltage inside. Now, from my last post, a new question cames to my head: If the voltage inside the cavity is constant, we have to expend no energy to move a free electron that were inside this cavity. Then, imagine you can "shrink" the sphere. As the inside voltage the electrons generate is constant, it will cost you nothing to move the electrons, but once you've reduced the sphere, the voltage is more negative. How this proccess happens? Will the electrons know that the voltage will decrease before it really decreases?. (Think about the graphic of potential respect the distance to the centre of the sphere, and you'll see that reducing the radius, the potential will decrease, but it stays always horizontal, so "in principle" it will cost you nothing to "shrink" the sphere). 



#2
May303, 06:44 AM

P: 353

Right ? 



#3
May303, 07:37 AM

P: 176

Yes, there is no current.
I'm talking about a 10 cm charged sphere with electrons. From the 10 cm to infinite, the voltage goes from V to 0 From the 10 cm to 0, the voltage remains at V Then, reducing the R of the sphere (imagine from 10 cm to 5 cm) cost you no energy, because the potential inside is constant, it doesn't matter the V is decreasing, it's constant on every value from R to 0. (you're not moving the electrons to a lower value of V, you are creating that V value by moving the electrons). Why should you expend work to reduce R, if the V inside the sphere remains constant on everypoint (but decreasing its value, it's true) on every moment?. 



#4
May303, 12:36 PM

P: 353

"compression" of an electrostatic charged sphere
What exactly do you mean that there is a voltage with infinity ?
As far as i remember we explain voltage as being an electrical potential energy between two charges. How do you make sure that there is a certain charge in infinity and that the charge on your sphere has a potential energy to the charge in infinity ? 



#5
May303, 03:30 PM

Sci Advisor
P: 1,341

Greetings !
Cala, I'm not certain what exactly you're talking about, but allow me to just make some general relevant clarifications. First, V is electric potential. What you may call voltage is U which is the difference in electric potential. Now, within a sphere U = 0 at every point, however the electric potenital V is the same at every point below or equal to R of the sphere. Now, if you decrease R  V will grow because V = K * Q / R . Now, since the potential is different when R is smaller (V is larger) it means that to achieve this radius decrease you need to put in energy which can be calculated by multiplying the potential difference U = V2  V1 and multiplying it by the charge of the sphere, so E = Q * U . (Think of it, you're pushing opposing charges closer together by overcoming their mutual electric repulsion.) Hope this helps. Live long and prosper. 



#6
May403, 05:34 AM

P: 176

Thanks Drag, you helped a lot.
You just answer exactly what i was talking about. So you've got to waste energy as if you were moving this charges from initial V to final V, although the V is created by the same charges, and not yet there. Thanks. 



#7
May403, 04:01 PM

Sci Advisor
P: 1,341





#8
May503, 05:59 AM

P: 176

To compute what energy you have to apply when you move a charge, you take the initial point, then look for the V at that point, then look at the final point where you'll move the charges, and there is another V value, and then make the calculus.
In the situation I explain, if you make the calculus by this method, the final V is not the V that is on final point with the charges on initial point. To make the calculus, you take the final V that will be there when the charges arrives there, not the V that is on final point when the charges are still on initial point (because the final V with charges on initial point will be the same as on the initial point, because the voltage inside the cavity is V elsewere). I hope it helps. 


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