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?
Yes that's correct. 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
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.
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
Is the absolute value of kynetic energy not relevant?, any speed will do, as long as it remains constant? In many textbooks I read slowly What is slowly?
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.
My point is that (KE), velocity cannot be left vague. 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?
If you travel from infinity with any finite speed, the time is infinite. If you half the speed, time remains the same, namely, infinite.
That's my quote 1 and 2: how do we calculate E-PE 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?
You are doing a mistake here, the work done on the particle doesnt depend on its velocity neither on the time it takes, because the force that does this work doesnt depend on velocity(the electric field is only due to the given charge which is considered stationary, while we consider the field from the moving unit charge negligible) neither on time (it is constant with time)
I do not know about v, that's why I was asking, but I surely know that the longer you expose anything to a force the greater is the KE acquired. Think about gravity!
If one says that PE is calculated bringing a test charge from infinity to anywhere and the time required to do this is infinite, then he is saying simply we cannot calculate it, at any speed. Am I wrong?
Yes. The work done by a conservative force is equal to the difference between the initial potential energy and the final potential energy: [tex] W_{\mathrm{cons}} = (E_{p})_{i} - (E_{p})_{f} [/tex] The work-energy 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 work-energy 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.
Potential energy in any force system is independent of time and dependant solely on position. 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 https://www.physicsforums.com/showthread.php?t=489731&highlight=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.
This is the maths formulation alright, and so I suppose it applies also to gravitation. 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?
thank you for the link. but the point is: is the definition given in the thread ,(you said it is correct), just a formal definition, meaning nothing in concrete.? or we can really use a test charge to measure Δ PE? or if we moved the two conductors of a capacitor? In a cyclotron you have a set difference in E-PE, and every time a charge makes a jump from a Dee it gets the same amount of KE, independently of its velocity. Is that correct?
It is a formal definition. Potential energy is not operationally defined physical quantity, but a derived quantity.