Electrostatic Potential Conceptby thebiggerbang Tags: acceleration, electric potential, kinetic energy 

#1
Aug711, 03:02 AM

P: 68

My textbook says that Electrostatic potential is the work done on a unit charge to bring it from infinity to a point from a given charge without acceleration, against the electric field presend due to the given charge.
So, as it says that there will be no acceleration, does it imply that there will be no change in the velocity of the charge that we are moving? And thus, does this imply that there is no change in the kinetic energy of the charge? 



#2
Aug711, 05:27 AM

P: 5,462

Another way to put this is to say that for its entire journey from infinity to its location, the charge is in equlibrium or passes through a series of equilibrium states. go well 



#3
Aug711, 06:02 AM

P: 3,015

Also, when a charge is accelerated, it emits electromagnetic radiation. It implies that no work is wasted in generating these propagating modes of the electromagnetic field.




#4
Aug711, 06:07 AM

P: 3,015

Electrostatic Potential Concept
BTW, this definition assumes that the electric potential at infinity is taken as zero (electric potential, like potential energy is determined up to an additive constant). This is possibly only for fields that decay fast enough at infinity, like, for example, the field of a point charge (that decays inversely proportional to the square of the distance). However, the field from a uniformly charged infinite line decays inversely proportional to the distance from the line and the potential is logarithmically divergent. The field of a uniform electric field gives a linearly divergent potential. Although all fields generated by real bodies decay sufficiently fast, sometimes it makes sense to take into account such idealized field sources for which your definition is not directly applicablel




#5
Aug711, 06:24 AM

P: 27

In many textbooks I read slowly What is slowly? 



#6
Aug711, 07:16 AM

P: 10

This is because you are dealing with electrostatics. If the charge is moving rapidly, you are dealing with electrodynamics and magnetic fields. I don't really know what exactly is slow enough to be described by electrostatic theorems, though.




#7
Aug711, 07:37 AM

P: 27

because if v is half, then time t is double. Then also total work done on charge (PE) is double, or more So what is the correct Electrostatic Potential Energy? 



#8
Aug711, 07:43 AM

P: 3,015

If you travel from infinity with any finite speed, the time is infinite. If you half the speed, time remains the same, namely, infinite.




#9
Aug711, 08:04 AM

P: 27

it seems we cannot calculate it! b)And what if you don't start from infinity? 



#10
Aug711, 08:08 AM

P: 3,015





#11
Aug711, 08:25 AM

P: 27

how do we calculate EPE from infinity to a point if time is infinite? b) if we do not start from infinity, the absolute value of velocity is not relevant? again, what is the threshold? 



#12
Aug711, 08:27 AM

P: 256





#13
Aug711, 08:30 AM

P: 3,015





#14
Aug711, 08:32 AM

P: 27

but I surely know that the longer you expose anything to a force the greater is the KE acquired. Think about gravity! 



#15
Aug711, 08:36 AM

P: 27

the time required to do this is infinite, then he is saying simply we cannot calculate it, at any speed. Am I wrong? 



#16
Aug711, 08:44 AM

P: 3,015

[tex] W_{\mathrm{cons}} = (E_{p})_{i}  (E_{p})_{f} [/tex] The workenergy theroem tells us that the total work done on an object is equal to the change in kinetic energy. In our case, since there is no acceleration, the velocity of the object remains the same, therefore the change in kinetic energy is zero, regardless of the speed of the object. Furthermore, there are two forces acting on the object at any time. The electrostatic force (no Lorentz force since the speed of the object is infinitely small) and the external force that counters it. Therefore, the workenergy theorem gives: [tex] W_{\mathrm{ext}} + W_{\mathrm{cons}} = 0 [/tex] Substituting for the work done by the conservative (electrostatic) force: [tex] W_{\mathrm{ext}} + \left((E_{p})_{i}  (E_{p})_{f}\right) = 0 [/tex] Solving for the final potential energy: [tex] (E_{p})_{f} = (E_{p})_{i} + W_{\mathrm{ext}} [/tex] Now, we choose the normalization that [itex](E_{p})_{i} = (E_{p})_{\infty} = 0[/itex]. Then: [tex] (E_{p})_{f} = W_{\mathrm{ext}} [/tex] This is the mathematical formulation of the sentence stated in the OP. 



#17
Aug711, 08:59 AM

P: 5,462

That includes gravity. We are discussing electric potential enrgy here. The definition is specifically worded to run from infinity to the position because we cannot calculate the work of separating charges. The calculation is easy and shown on post 40 of this thread http://www.physicsforums.com/showthr...ight=potential Your book says 'slowly' instead of 'equilibrium' or 'without acceleration' I expect. They are all equivalent statements. The represent the desired condition that as the charge is brought in from infinity none of the work done against the electric force is converted to kinetic energy  it is all stored as electric potential energy. This condition gives validity to the calculation described above. 



#18
Aug711, 09:02 AM

P: 27

and I suppose it applies also in the case we do not start from infinity. (point b) Imagine we have in vacuum 55200 esu,0.0000148 C ,positive charge ( 4x 10^14 at 1m). An electron is at rest at r0= 10^7m/ (^9 cm.) (acc= 4 m/sec^2, 400cm/s^2). If we move it to r = 6.4x 10^6m/ (^8cm) (acc= 9.8 m/s^2 ,980cm/s^2) is work done on the charge the same if v changes? or is it useless and we need only Maths and KE= Δ PE = 2.25 x 10^ 7 ? That is to say the same KE an electron would get anyway in a free fall from r0 to r? 


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