Physics Asked on January 13, 2021
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 = Ccdot V $ as $C$ and $V$ remain the same. The charge on each plate comes out to be $CV$.
Is the formula $C = Qcdot V$ still applicable or not and what will the charge distribution be like?
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.
Correct answer by Anonymous on January 13, 2021
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.
Answered by Bob D on January 13, 2021
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