Capacitors and Capacitance, Combination of Capacitors in Series and Parallel

Class XI covered the RESISTOR: a component that allows current to pass through it, but which only ever DISSIPATES the associated electrical energy as heat, with no capacity to store it. A genuinely different kind of device is needed for a circuit to STORE electrical energy for later use -- and the most common such arrangement is a pair of conducting plates carrying equal and opposite charges, separated by some dielectric medium (which may simply be air, or vacuum).

For two conductors 1 and 2, carrying charges $+Q$ and $-Q$ respectively, with potential difference $V=V_1-V_2$ between them, the field in the region between them -- and hence the potential difference itself -- is always directly proportional to the magnitude of the charge $Q$ placed on them. Since $V\propto Q$, the ratio $Q/V$ is therefore a CONSTANT for any given pair of conductors, called the CAPACITANCE, $C=Q/V$: a purely geometric property, depending only on the size, shape, separation and any intervening dielectric of the two-conductor system -- never on $Q$ or $V$ individually, which can each be varied freely while their ratio stays fixed.

The SI unit of capacitance is the FARAD (F), where $1\,\text{farad}=1\,\text{coulomb}/1\,\text{volt}$: a capacitor is said to have a capacitance of exactly one farad if its potential difference rises by exactly 1 volt for every 1 coulomb of charge given to it. In practice, the farad turns out to be an enormous, almost never directly usable unit, so real capacitance values are almost always quoted in its much smaller sub-multiples: $1\,\mu F=10^{-6}$ F, $1\,\text{nF}=10^{-9}$ F, and $1\,\text{pF}=10^{-12}$ F.

The underlying PRINCIPLE of how any capacitor manages to store more charge at a given voltage than an isolated conductor could is worth working through explicitly. Take an isolated, charged metal plate $P_1$ (area A, charge $+Q$, potential V, so its own bare capacitance is $C_1=Q/V$). Now bring a second, initially uncharged, INSULATED metal plate $P_2$ near it (not touching). By electrostatic INDUCTION, a negative charge is induced on $P_2$'s face nearer $P_1$, and an equal positive charge appears on $P_2$'s farther face. Because the induced negative charge on the near face is CLOSER to $P_1$, it has a stronger, more effective influence than the farther, weaker positive charge -- so on balance, $P_1$'s own potential is LOWERED by $P_2$'s mere presence. If $P_2$'s outer (far) face is now connected to EARTH, its induced positive charge -- being genuinely free and mobile -- simply drains away into the ground, leaving only the induced negative charge bound on $P_2$'s near face (held in place purely by its attraction to $P_1$'s own positive charge); this earthing step lowers $P_1$'s net potential difference still further. Since capacitance is charge divided by potential difference, and the charge $Q$ on $P_1$ has not changed at all while its effective potential difference has been driven down, the two-plate arrangement's capacitance must be GREATER than $P_1$'s own isolated $C_1$: bringing an earthed conductor near a charged one always raises the overall system's capacitance. This is the general PRINCIPLE common to every capacitor, whatever its specific geometry. …