User:Inconspicuum/Physics (A Level)/Capacitors

If you place two conducting plates near each other, with an insulator (known as a dielectric) in between, and you charge one plate positively and the other negatively, there will be a uniform electric field between them. Since:

$$E = \frac{V}{d}$$,

as the distance between the two plates decreases, the energy stored increases. This system is known as a capacitor - it has a capacitance for storing charge. The capacitance C of a capacitor is:

$$C = \frac{Q}{V}$$,

where Q is the charge stored by the capacitor, and V is the potential difference between the plates. C is therefore the amount of charge stored on the capacitor per unit potential difference. Capacitance is measured in farads (F). Just as 1 coulomb is a massive amount of charge, a 1F capacitor stores a lot of charge per. volt.

Any capacitor, unless it is physically altered, has a constant capacitance. If it is left uncharged, Q = 0, and so the potential difference across it is 0. If a DC power source is connected to the capacitor, we create a voltage across the capacitor, causing electrons to move around the circuit. This creates a charge on the capacitor equal to CV. If we then disconnect the power source, the charge remains there since it has nowhere to go. The potential difference across the capacitor causes the charge to 'want' to cross the dielectric, creating a spark. However, until the voltage between the plates reaches a certain level (the breakdown voltage of the capacitor), it cannot do this. So, the charge is stored.

If charge is stored, it can also be released by reconnecting the circuit. If we were to connect a wire of negligible resistance to both ends of the capacitor, all the charge would flow back to where it came from, and so the charge on the capacitor would again, almost instantaneously, be 0. If, however, we put a resistor (or another component with a resistance) in series with the capacitor, the flow of charge (current) is slowed, and so the charge on the capacitor does not become 0 instantly. Instead, we can use the charge to power a component, such as a camera flash.

Exponential Decay
Current is the rate of flow of charge. However, current is given by the formula:

$$I = \frac{V}{R}$$

But, in a capacitor, the voltage depends on the amount of charge left in the capacitor, and so the current is a function of the charge left on the capacitor. The rate of change of charge depends on the value of the charge itself. And so, we should expect to find an exponential relationship:

$$Q = Q_0e^{-\frac{t}{RC}}$$,

where R is the resistance of the resistor in series with the capacitor, Q is the charge on the capacitor at a time t and Q0 was the charge on the capacitor at t = 0. Since Q = I&Delta;t:

$$I\Delta t = I_0\Delta te^{-\frac{t}{RC}}$$

$$I = I_0e^{-\frac{t}{RC}}$$,

where I is the current flowing at a time t and I0 was the initial current flowing at t = 0. Since V = IR:

$$V = V_0e^{-\frac{t}{RC}}$$

The power being dissipated across the resitors in the circuit is IV, so:

$$P = I_0V_0e^{-\frac{t}{RC}}e^{-\frac{t}{RC}} = P_0e^{-\frac{2t}{RC}}$$

Energy
The energy stored by a capacitor E is defined as:

$$E = \int^V_0 Q \; dV$$

In other words, it is the area under a graph of charge against potential difference. Charge is proportional to potential difference (Q = CV), so the area under the graph is that of a triangle with base V and height Q. You can show this mathematically:

$$E = \int^V_0 CV \; dV = C\left [\frac{V^2}{2}\right ]^V_0 = \frac{1}{2}CV^2$$

Since Q = CV:

$$E = \frac{1}{2}QV$$

Circuits
The circuit symbol for a capacitor is. A simple circuit with a capacitor in series with a resistor, an ideal ammeter (no resistance), and in parallel with an ideal voltmeter (infinite resistance) looks like the following:



In the position shown, the capacitor is charging. If the switch were put in the other position, the capacitor would be discharging exponentially through the resistor. In this circuit, the capacitor charges instantly since there is no resistance to slow it down. In reality there will be internal resistance in the battery, meaning that the capacitor charges exponentially.

If capacitors are placed in parallel, they act as one capacitor with a capacitance equal to the total of all the capacitances of all the individual capacitors. If capacitors are placed in series, the distances between the plates in each of them result in the capacitance of the imaginary resultant capacitor &Sigma;C being given by:

$$\frac{1}{\Sigma C} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}$$