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How is the charge +q given to the one plate to the capacitor with the help of a battery, does the battery remove electrons from the plate and +ve charge gets accumulated on one plate?Does it?

And one plate gets charge induction of other plate occurs which creates potential difference between the two plates and then the +ve charge of the 2nd plate is grounded and passes through to the ground.

Is this process correct?

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  • $\begingroup$ You are right, the positive terminal of the battery removes electrons from the plate. The other plate needs to be to the negative of the battery to accept electrons. If the negative plate is grounded or not, it will not make a difference in a simple battery-capacitor circuit. $\endgroup$
    – CAGT
    Commented Feb 1, 2014 at 2:48

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When the battery is first connected to the capacitor, a current does flow. This current removes electrons from one side of the capacitor, and deposits an excess of electrons on the other side. As this happens, the voltage between the plates of the capacitor rises. When it is equal to the battery voltage, no more current can flow and the capacitor is fully charged.

In practice the capacitor's voltage will always be lower that the battery but after a short time the difference will be immeasurably small. The voltage on the capacitor rises as given by the equation $$V(t) = V_0(1-e^{-t/ \tau})$$ If you plot this function you'll see that the voltage approaches the battery voltage ($V_0$) asymptotically: it gets ever closer but never quite reaches it.

In the equation, $\tau$ is the time constant of the circuit.$$\tau=RC$$

Here $C$ is the capacitance (in Farads) and $R$ is the resistance of the circuit, in ohms. This resistance includes the internal resistances of the battery and the capacitor, the resistance of the wires and any other resistors you may have put between the battery and capacitor.

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Disregarding conventional current. Whenever a capacitor is connected to a voltage source (battery) the electrons flow from high potential to low, now being that the capacitor is in series with the battery the electrons will start to pile on one side of the plate.

But by doing this since the electrons have a natural electric field to them the electric field created by the electrons permeates through the dielectric repelling its electrons which in turn repel the electrons from the other plate but as these valence electrons are repelled from the plate and start a chain reaction that allows current to flow for a short time the positive nuclei that have been stripped of the valence electrons start to gather on the same plate since they are attracted to the electric field created by the electrons.

So as one plate is saturated on one side with electrons and the other with positive nuclei it reaches a point where there is a very high concentration and the current is insignificant, at this point the capacitor is said to be charged.

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  • $\begingroup$ That middle run-on-sentence has 92 words in it. Please break you thoughts down into more concise and easily digestible chunks. $\endgroup$ Commented Aug 2, 2014 at 18:53
  • $\begingroup$ Sometimes you've got to let it flow..... $\endgroup$
    – AlanZ2223
    Commented Aug 2, 2014 at 19:00

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