The capacitance of a parallel plate capacitor is given by C=εr A/d, where A is the area of the plates, d is the separation between the plates and εr is the relative permiliability of the dielectric between the plates. The relative permiability is some factor, K multiplying the permiability of free-space ε0. ε0 has a value of 8.85Ч10-12 F.m-1.
A full list of relative permiabilities can be found for almost any dielectric material. The greater the relative permiability the greater the capacitance of the capacitor. Some good materials are Mica, polystyrene, oil.
εr=K ε0
Capacitor Networks
Figure 3. Capacitor networks in Series and Parallel
Series
Consider the series network of capacitors shown in Figure 3a. where the positve plate is conected to the negative plate of the next.What is the equivalent capacitance of the network? Look at the plates in the middle, these plates are physically disconected from the circuit so the total charge on them must remain constant. It follows that when a voltage is applied across both of the capacitors, the charge + Q on the positive plate of capacitor C 1 must be balanced by the charge – Q on the negative plate of capacitor C 2. The net result is that both capacitors possess the same charge Q. The potential drops V 1 and V 2 across the two capacitors are in general, different. However, the sum of these drops equals the total potential drop V applied across the input and output wires. V = V 1 + V 2. The equivalent capacitance of the pair is again C T= Q / V. Thus, 1/ C T = V / Q = (V 1 + V 2)/ Q = V 1/ Q + V 2/ Q giving;
In general, for N capacitors connected in series, is
By connecting capacitors in seires you store less charge so does ever make sense to connect capacitors in series? It is sometimes done because capacitors have maximum working voltages, and with two 900 volt maximum capacitors in series, you can increase the working voltage to 1800 volts.
