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What would happen to capacitance and other quantities like Voltage if I initially take a Parallel Plate capacitor with each plate having equal charge "$q$" and then connect it to a constant voltage supply (battery)?

I know Capacitance is a geometrical quantity and the battery doesn't supply any net charge to capacitor plates. But if we go by this definition, we know that across the Capacitor plates, the voltage drop = EMF of battery (ignoring internal resistance) and if we go by the relation $Q = C\cdot V $ as $C$ and $V$ remain the same. The charge on each plate comes out to be $CV$.

Is the formula $C = Q\cdot V$ still applicable or not and what will the charge distribution be like?

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  • $\begingroup$ @PranavAggarwal So You are saying. q ( charge already present )+ Q ( charhe Given by battery) = C*V. ? If that is the case then net charge on both plates becomes zero .which is not possible here .? $\endgroup$
    – Normalitee
    Commented Nov 22, 2020 at 7:59
  • $\begingroup$ The last equation C=Q.V is incorrect. It should be C=Q/V. Assume its a type $\endgroup$
    – Bob D
    Commented Nov 22, 2020 at 15:17

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Assuming that you gave charge of equal magnitudes and signs on the plate. So initially the charges would redistribute themselves so as to nullify the electric field inside the conductor. When you will do calculations you would find out is that the charges distribute on the outer surfaces of the plates.

Hence giving an initial potential to the plates but zero potential difference. As you will connect the battery. The battery supplies the charges and hence charging the capacitor as usual and the charges outside would remain unaffected.

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  • $\begingroup$ Thanks Bro ~ I reached the same conclusion after brainstorming and verifying it with you now i am sure its correct. $\endgroup$
    – Normalitee
    Commented Nov 22, 2020 at 9:13
  • $\begingroup$ Can you clarify the statement "Hence giving an initial potential to the plates but zero potential difference"? $\endgroup$
    – Bob D
    Commented Nov 22, 2020 at 15:24
  • $\begingroup$ @BobD initial potential as the plate becomes charged and charge is distributed on the outer surfaces. The sheet acquires a potential that means we have todo work in bringing the chargefrom infinity to that point but both the plates acquires potential equally. Hence no new potential difference is generated. $\endgroup$
    – Anonymous
    Commented Nov 22, 2020 at 16:17
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In the equation

$$C=\frac{Q}{V}$$

The capacitance $C$ is a constant because it is based on the physical characteristics of the device, namely the plate area, separation distance and the dielectric of the material between the plates in the case of a parallel plate capacitor.

The charge $Q$ is the net charge on each plate, positive on one plate and an equal amount of negative charge on the other plate.

The voltage $V$ is the potential difference between the plates due to the net charge on each plate.

A battery does not supply charge to the capacitor. If its Emf is greater than the capacitor voltage and polarity the same, it will do work to remove electrons from the positively charged plate and deposit an equal amount of electrons on the negatively charged plate thereby increasing the net charge on each plate. It will do this until the voltage between the plates equals the battery Emf.

Hope this helps.

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  • $\begingroup$ Why the anonymous downvote? $\endgroup$
    – relayman357
    Commented Nov 22, 2020 at 14:20
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    $\begingroup$ @relayman357 I wondered that too. I see nothing wrong with my answer. Sometimes an answer is down voted because it is a solution to a homework and exercise question. But I feel mine was simply addressing conceptual issues on the part of the OP. $\endgroup$
    – Bob D
    Commented Nov 22, 2020 at 15:06
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    $\begingroup$ @relayman357 Though there is nothing wrong with down voting an answer anonymously, it is helpful to understand the reason and, if justified, learn from it. $\endgroup$
    – Bob D
    Commented Nov 22, 2020 at 15:08
  • $\begingroup$ Yes, I earned one of those legitimate down votes just this morning myself. :-). But, in case’s like your answer above, I don’t understand the downvote without explanation of what is wrong. If you missed something, or mis-stated it, you should be given the opportunity to correct it and have the downvote removed. $\endgroup$
    – relayman357
    Commented Nov 22, 2020 at 15:09
  • $\begingroup$ @relayman357 BTW what answer of yours was downvoted? $\endgroup$
    – Bob D
    Commented Nov 22, 2020 at 17:31

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