# Deriving the Exponential Capacitor Decay equation

If you’ve studied capacitors, you’ll have (very probably) come across the decay equation:
$Q = Q_0 e^\frac{-t}{RC}$ where $R$ is resistance, $C$ is capacitance, $Q$ is the charge on the capacitor at time $t$ and $Q_0$ is the initial charge.

#### Where does this come from?

Firstly from the basics of circuits, we know that $V = IR$ (yes, I’m using the british convention with regards the symbol for voltage, I understand elsewhere $U$ is used instead).
$V = IR \Rightarrow I = \frac{V}{R}$

Now when we discharge a capacitor through a circuit, the e.m.f.s are supplied by the capacitor.
$Q = CV \Rightarrow V = \frac{Q}{C}$

Now substitutiting for V in the above equation gives
$I = \frac{V}{R} \Rightarrow I = \frac{\frac{Q}{C}}{R} = \frac{Q}{RC}$

So we now have an equation for the current in the circuit, but wait, where does this current come from? It’s the capacitor discharging – this current is equal to the rate of discharge of the capacitor, so:
$-I = \frac{{\mathrm d}Q}{{\mathrm d}t}$
So substituting the above expression for I
$-\frac{Q}{RC} = \frac{{\mathrm d}Q}{{\mathrm d}t}$

Ooh look, a differential equation.
Seperate the variables giving:
$\int -\frac{1}{RC} \, {\mathrm d}t = \int \frac{1}{Q} \, {\mathrm d}Q$
$\Rightarrow \frac{-t}{RC} = ln(Q) + ln(k)$ where k is some constant.
$\Rightarrow e^\frac{-t}{RC} = kQ$
$\Rightarrow Q = \frac{1}{k}e^\frac{-t}{RC}$

If we define the initial charge to be $Q_0$ – the charge when t = 0
$\mathrm{ let } t = 0 \Rightarrow Q_0 = \frac{1}{k}$
so we now can replace $\frac{1}{k}$ giving:

$Q = Q_0 e^\frac{-t}{RC}$

MY WORK HERE IS DONE

:D