Electric potential and increased capacitance I could really do with some insight into how the capacitance of a conductor increases when you bring another oppositely charged or earthed conductor near it. The explanation in my book is as follows:


What I can't understand is how the potential ($V$) is decreased. If I bring an oppositely charged conductor near the conductor than as the distance decreases the electric force increases and $V = k_0 \frac{q}{r}$ so $V$ must increase also. I know that I am just not seeing what is happening clearly, but everywhere I look for an answer someone says $C = \frac{Q}{V}$ so if $V$ decreases $C$ must increase which I don't find very helpful.
 A: (a) $=_0 \frac qr$ applies only when $r$ is so far from the charged plate that the plate behaves as a point charge. Near the plate ($r<<$ plate length) the potential varies linearly with $r$. It is therefore not easy to give a formula for the potential near the plate using the usual convention of zero potential at infinity (except in the case of a charged disc, along its axis). All the same we can argue qualitatively...
(b) Potentials add as scalars. The potential at a distance $r$ from the positive plate will therefore be reduced when we place the negative plate as shown, because the negative plate will contribute a negative potential at this point. [Equivalently, the amount of work we have to do to per unit charge to take a testing charge, $q$, from infinity to the point is reduced, because the negative plate is pulling $q$.
(c) I'm afraid that the last step in the argument has to be the one you don't like: the appeal to $C=\frac QV$, telling us that, for a given charge, $Q$ on the plate, if $V$ goes down, $C$ goes up. But $C=\frac QV$ is inescapable, because it is the very definition of capacitance.
A: Diagram b is misleading in that if the magnitude of the negative charge on the right-hand plate is the same as magnitude of the positive charge on the left-hand plate then almost all of the positive charge and negative charge will reside on the sides of the plates facing each other.
This means that the electric field outside the volume between the two plates is very small compared with the electric field around the positively charged plate in diagram a.
Thus the work done in bringing a positive test charge up the the positively charged plate is reduced.
A similar thing happens when an earthed plate is brought close to the positively charged plate.
The induced negative charged are attracted to the positive plate and the induced negative charge run away to earth.
Again the electric field outside the volume between the two plates is reduced and so the work done in bringing a positive test charge up the the positively charged plate is reduced.
