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I was looking for an intuitive explanation for the capacitance between two plates, couldn't find any suitable though. So I tried to figure it out myself and I am wondering if it is correct.

Capacitance is defined as the ratio of the charge(Q) of one of the two plates and the potential difference(V) across the two plates (after they have been charged, of course) and we know that P.D (V) means that the amount of work needed to bring one electron from one point to another. So when we divide the number of charges by the P.D, what we are determining is that how many charges can possibly go from one plate to another by consuming that PD. After they have switched the sides the PD will again get to zero. for example (it is just an supposition, don't take it seriously) V = 2 v and Q = 6 then C = 3 c/v which means that the plates can accumulate 3 coulomb of charges. As we increase the V, Q increases too since V is proportional to Q/t and the ratio never differs unless we change the physique or vicinity of the plates.

Am I right with my explanation ? if not, Where did I go wrong ? Any kind of help or inquiry is appreciated.

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Let's take it one point at a time.

Capacitance is defined as the ratio of the charge(Q) of one of the two plates and the potential difference(V) across the two plates (after they have been charged, of course)

Capacitance is the amount of charge that can be stored on the plates per volt across the plates, or

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

The statement "after they have been charged, of course" is not essential to the definition of capacitance. This is because capacitance is also defined as

$$C=\frac{εA}{d}$$

where $ε$ is the permittivity of the space between the plates, $A$ is the area of each plate and $d$ is the separation between the plates. Note that voltage is not part of this definition.

and we know that P.D (V) means that the amount of work needed to bring one electron from one point to another.

Actually, PD is defined as the work required, per unit charge, to move the charge between two points. For a capacitor with constant field $E$ between the plates, the force exerted on a charge between the plates is $QE$ and the work done to move the charge between two points separated by a distance $d$ is $QEd$. Therefore the PD is $Ed$.

So when we divide the number of charges by the P.D, what we are determining is that how many charges can possibly go from one plate to another by consuming that PD.

No. Dividing the number of charges by the PD, or $\frac{Q}{Ed}$ makes no sense. Now if you multiplied some total amount of charge times the PD that would give you the total work required to move all those charges a distance $d$. You don't consume PD. You consume energy moving the charges. For a capacitor connected to a battery, the battery supplies that energy.

After they have switched the sides the PD will again get to zero. for example (it is just an supposition, don't take it seriously) V = 2 v and Q = 6 then C = 3 c/v which means that the plates can accumulate 3 coulomb of charges. As we increase the V, Q increases too since V = QR/t (R = resistance) and the ratio never differs unless we change the physique or vicinity of the plates.

It's difficult to address all of this since it is based on an erroneous understanding of capacitance and PD, which I have already addressed. Please look over my answer and then let me know if your last statement still makes sense to you.

Hope this helps.

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  • $\begingroup$ Referring to the third part of your explanation, I didn't s say dividing Pd by Q. Instead it was the other way around. Talking about the second part, I admit I was wrong about moving electrons because it is actually the work done per unit positive charge which while being moved across two points. Nonetheless, I don't quite get the point you are trying to make there with Q, E and d. It makes no sense to me atleast for now. $\endgroup$
    – Hardik Sharma
    Commented Jun 16, 2019 at 16:45
  • $\begingroup$ Also by consuming PD I meant consuming the work (or specifically energy) which is responsible for moving electrons ( or current), if that makes sense. Correct me if I am wrong because it is getting a bit confusing since electrons do not actually move from one plate to another while that is being charged, do they ? Also , as I said, energy is required not provided that might get consumed. $\endgroup$
    – Hardik Sharma
    Commented Jun 16, 2019 at 16:53
  • $\begingroup$ Or may be it is just an estimation . ( Movement of the net charge by the influence of PD. Net charge -> total charge that was stored on one plate before it moved to another when the plates were connected.) $\endgroup$
    – Hardik Sharma
    Commented Jun 16, 2019 at 16:57
  • $\begingroup$ Plate Regarding your first comment I realized that and corrected it. Regarding second comment PD is not work. It’s work per unit charge. You are correct electrons don’t move from plate to plate. It’s the work required by the battery to move electrons from the positive plate to the negative plate. Actually by convention it’s the work required by the battery to move positive charge from the negative to the positive Plate $\endgroup$
    – Bob D
    Commented Jun 16, 2019 at 17:04
  • $\begingroup$ You didn’t give me enough to answer the second comment. See above $\endgroup$
    – Bob D
    Commented Jun 16, 2019 at 17:07

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