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Point charge ; current in a circuit

  1. Oct 25, 2013 #1
    Hello , there are two questions I would like to ask from a educational paper i just read.

    For the first question please look the attached picture , it says that in the picture of the capacitor with the wire that has a gap in the middle and next to it when the gap is closed , it says that in the exact middle of that wire between the two plates the wire is neutral , that would imply no charge , well first of all even if that happens , for how long as i basically see a charged capacitor being short circuited with a wire which would basically discharge the cap.

    Let me rephrase the question , imagine a LC parallel tank circuit , only in the middle of the inductor cut the wire in half and place a switch in series , now charge the capacitor while the switch is open , once the capacitor is charged , close the switch , the question is for the first instant would the switch being in the middle feel any current/voltage through it ? as the current would form when the switch is closed but because going through an inductor it would build up a magnetic field which would oppose the current so at the first instant the inductor would act as a " choke" , so what happens with the switch in the middle ?

    there has been some discussion about this here on PF but I still haven't got a clear answer so I'm going to ask this one more time.
    The paper i read says that the electric potential is independent of the value of q ,(q being the fundamental charge strength of a charged particle)
    So does that mean that I can have one electron at say million volts of potential which has the same field strength around it as say 10 electrons at the same potential or no?

    Attached Files:

  2. jcsd
  3. Oct 26, 2013 #2


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    Staff: Mentor

    The full voltage drop of the circuit is felt across the open switch initially. After closing the switch current will flow and the voltage drop will be divided appropriately between all components in the circuit. I don't know about "the first instant".

    If we assume the switch has identical resistance and is the same dimensions as the length of wire it replaced, then the inductor acts the same in both cases.

    I'm not sure. Wiki says the following about electric potential:

    The electric potential at a point is equal to the electric potential energy (measured in joules) of any charged particle at that location divided by the charge (measured in coulombs) of the particle. Since the charge of the test particle has been divided out, the electric potential is a "property" related only to the electric field itself and not the test particle. The electric potential can be calculated at a point in either a static (time-invariant) electric field or in a dynamic (varying with time) electric field at a specific time, and has the units of joules per coulomb (J C–1), or volts (V).
  4. Oct 26, 2013 #3
    Well I guess very little to no current flows through the switch in the first instant (by instant I mean the time it takes for a certain inductor , voltage and current to produce an magnetic field in the coil)

    So after you close the switch there should be a small time period at which all or most of the inrushing current is blocked due to the back emf of the coil right? After the field decreases the current starts to run but what happens at the moment and shortly before the moment of saturation that I guess is the question? As to my mind the current should be busy building up the field.

    Still confused about the charged particle thing , well it says that " The electric potential at a point is equal to the electric potential energy (measured in joules) of any charged particle at that location divided by the charge (measured in coulombs) of the particle." As long as this goes it kinda makes me think that the potential is the overall potential divided by the particle potential , but that doesn't sound right.
    But what if I have a gazillion volt potential over some physical are "x" and this area happens to be only one electron? What can I divide then?
  5. Oct 26, 2013 #4


    Staff: Mentor

    The solution to a LC circuit is given here:

    You start out with a voltage across the capacitor and no current. The current increases and the voltage decreases and they oscillate in a sinusoidal pattern.

    That means that the electric potential does not depend on whether the charge on an electron is 1.6E-19 C or 2.8E-19C or 1.1E-19 C or whatever. All that changes is how many electrons are required to get the same charge.
  6. Oct 26, 2013 #5


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    Staff: Mentor

    The strength of the magnetic field is irrelevant. It is the rate of change in the field that causes back EMF. That's why, initially, the current is almost totally blocked. The rate of change is very very high even though the field is very small.
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