# Taha Azzaoui - 2018.07.10

### Parallel Plate Capacitors

Often times we wish to store electrical energy for later use. To do so, we can use a parallel plate capacitor, an electrical component which stores potential energy in the form of an electric field. If we connect a battery to both plates of the capacitor, the battery will transport charge from one plate to the other until the potential difference within the plates is equivalent to that of the battery. To analyze the behavior of capacitors, we use the quantity of capacitance, which is defined as the ratio of the magnitude of the charge on each plate, Q, and the potential difference across the plates ΔVc. That is,
$$C = \frac{Q}{\Delta V_c} \tag{1}$$
Note that the SI unit of capacitance is Coulombs per volt, otherwise known as a Farad.

### The RC Circuit

An RC (short for resistor-capacitor) circuit is an aptly named circuit consisting of a resistor of resistance R, a capacitor of capacitance C, and a battery of electromotive force ε. Let Q be the magnitude of the charge on the capacitor and let I be the magnitude of the current flowing through the circuit. Note that these two quantities are related by
$$I = \frac{dQ}{dt} \tag{2}$$

### Capacitor Charging

When the voltage source (in this case a battery) is connected to the circuit, the current flowing through the circuit will serve to charge the capacitor until its potential difference is equivalent to that of the battery. According to Kirchhoff’s second law, we much have $\sum_{i = 1}^{n}V_i = 0$. We can express the potential difference across the capacitor in terms of capacitance using (1), $V_c = \frac{Q}{C}$. Finally, we’ll assume that the potential difference across the battery is equivalent to its electromotive force. Putting all this together, we get the following first order linear differential equation.
$$R\frac{dQ}{dt} = \frac{Q}{C} - \varepsilon = 0 \nonumber \tag{3}$$
The equation here is separable, and can be solved trivially as follows
$$\begin{eqnarray} \frac{dQ}{dt} + \frac{1}{RC}Q - \frac{1}{RC} \varepsilon = 0 \nonumber\\\\ \frac{dQ}{dt} = \frac{Q - \varepsilon C}{-RC} \nonumber \\\\ \int \frac{1}{Q - \varepsilon C} dQ = \frac{-1}{RC} \int dt \nonumber \\\\ ln \| Q - \varepsilon C\| = \frac{-t}{RC} + K \nonumber \\\\ Q - \varepsilon C = e^{\frac{-t}{RC}}e^{K} \nonumber \\\\ Q = \varepsilon C + e^{\frac{-t}{RC}} e^{K} \nonumber \\\\ Q(0) = \varepsilon C + e^{K} = 0 \implies e^K = - \varepsilon C \nonumber \\\\ Q = \varepsilon C - \varepsilon C e^{-t}{RC} \nonumber \end{eqnarray}$$

Therefore, we have
$$Q(t) = \varepsilon C(1 - e^{-t}{RC}) \tag{4}$$
Note that
limt → ∞Q(t) = εC
Which is exactly what we would expect given the definition of capacitance in (1).

### Capacitor Discharging

We now turn to what happens when the voltage source is disconnected from the circuit. Our equation is the same as before, save for the electromotive force which has now disappeared.
$$R \frac{dQ}{dt} + \frac{Q}{C} = 0 \nonumber \tag{5}$$

$$\begin{eqnarray} \frac{dQ}{dt} + \frac{1}{RC}Q = 0 \nonumber \\\\\int \frac{1}{Q} dQ = \frac{-1}{RC} \int dt \nonumber \\\\ ln \|Q\| = \frac{-t}{RC} + K \nonumber \\\\Q = e^{-t}{RC}e^{K} \nonumber \\\\\text{Note: } Q(0) = e^0e^K \implies Q_0 = e^{K} \nonumber \\\\ \end{eqnarray}$$

Therefore, we have our desired result:
$$Q(t) = Q_0e^{\frac{-t}{RC}} \tag{6}$$