# Can I calculate the current of any setup with capacitance by the same formulae?

There are known formulae relating a capacitor's voltage and current in its classic form (battery, wires, a capacitor circuit), but what if we had a charged plate (-ve), then we put on one of its surfaces a dielectric, then we put another plate (neutral, connected to the ground)?

Technically, this is a capacitor (2 differently charged plates separated by a dielectric). The other plate will be +ve charged (like charging by induction) and a current will flow in the wire connecting the second plate to the ground. If I wanted to calculate this current, should I use the same formulae or will they change because the voltage is changing only on one plate and the second plate will reach a equal and opposite voltage pretty quickly?

Yes, you can use the standard formula. the equations \begin{equation} I = C \frac{\mathrm{d}V}{\mathrm{d}t} \end{equation} or equivalently \begin{equation} Q = C V \end{equation} are really just the definition of capacitance. (similarly the equations for a resistor and inductor are just definitions of resistance and inductance) Any real system will have some resistance, some capacitance and some inductance. The capacitors, resistors and inductors we talk about in circuit theory are just idealisations, but we can very often write real systems as a combination of these ideal components wired up in some way.

As for the system you have described, that should essentially act as two capacitors in series (you can imagine if you split the earthed plate in two with a wire between the two parts then in the gap between them there would be no electric field, so nothing would really have changed), so you can find the total capacitance from \begin{equation} \frac{1}{C_{total}} = \frac{1}{C_1}+\frac{1}{C_2} \end{equation} If you wanted to find the charges on each plate you would have to specify the potential differences between each of your batteries terminals and the earth, but if you are only interested in the currents in the circuit then you should be able to treat the combined capacitor as a black box with a certain capacitance and the usual p.d. accross it should be enough.

If you wanted to calculate the capacitance for some complicated setup in general, you can do this by specifying the charges on the various parts of your system, from that working out what the electric fields produced are, and then integrate the fields between your contact points to find the potential difference. You can then find the capacitance from $C = \frac{Q}{V}$. The linearity of Maxwell's equations mean that this should always come out as something that depends only on the geometry, and not on any electrical quantities, so provided that none of the plates move, no electrical contacts are connected or disconnected, etc the capacitance should be constant.

• you should consider the answer by @alfred centauri , the capacitance won't be constant , while assembling the apparatus, as the metal grounded plate approaches the dielectric, a charge is induced in it , as the distance decreases between then the charge and capacitance increases – Abdelrahman Esmat Aug 7 '14 at 7:20
• I don't think that's what is being asked about. I think read the question as simply saying you have three plates in succession, one of which is earthed. If you are adding the middle plate while the circuit is running then of course the capacitance will not be constant, as indeed I say in the last paragraph, but I don't think that is what is being asked. could @Alfred Centairi clarify? – By Symmetry Aug 7 '14 at 7:30

If I wanted to calculate this current, should I use the same formulae

No, since, for one, the capacitance $C$ of the apparatus changes as the capacitor is assembled. In the equation

$$i = C \frac{dv}{dt}$$

$C$ is a constant. There are other considerations too but the bottom line is, no, you cannot use the above formula to determine the current in the wire connecting the second plate to ground as the capacitor is assembled.

So we should put C as a function of time too (dC/dt) right?

When C is not constant, we have

$$i = C\frac{dv}{dt} + v\frac{dC}{dt}$$.

• So we should put C as a function of time too (dC/dt) right? – Abdelrahman Esmat Aug 7 '14 at 7:21
• @AbdelrahmanEsmat, correct. If C is not constant, we have $$i = C\frac{dv}{dt} + v\frac{dC}{dt}$$. – Alfred Centauri Aug 7 '14 at 11:51