# A-level Physics/Forces, Fields and Energy/Capacitors

Introduction When two conductive materials are separated by an insulating material, then it will behave as a Capacitor with associated Capacitance in the units of Farads (Coulombs/Volt).
Intuitively, Capacitance can be interpreted as "How much charge can I shove into a material if I apply a certain voltage?" Capacitors are useful because it can store energy momentarily and dissipate the energy later, and with combination of a resistor, it is capable of "delaying" a signal.

## Definition of Capacitance/Capacitor

A capacitor is usually made from two sheets of metal separated by an insulating material (such as air or ceramics). If we apply a voltage between the two sheets, there will be an associated electric field generated, and charges will accumulate on each side of the plates. We define Capacitance ($C$) to be $C={\frac {Q}{V}}$ where Q is the charge that accumulates on the plate when voltage V is applied. This can be concluded intuitively. A capacitor of impossibly high quality will be able to accumulate the maximum amount of charge with the least amount of voltage applied. The unit of capacitance is in Farad, or F for short.
Another definition for capacitance is area of the plates times the permittivity, divided by the distance of the two plates from the insulating material. In other words, $C={\frac {\epsilon A}{D}}$. This makes intuitive sense - if we make the plates bigger, we can store more charge, and if we bring the plates closer, the tendency for the charges to attract increase, thereby increasing the electric field generated. Further, because of the nature of a capacitor, as the ability for the dielectric (in between the two plates) to restrict electrical fields (permittivity) goes up, so does the capacitance,
Now, it does not mean that capacitance is a property that appears only on two sheets of metallic sheet. In fact, any piece of wire or metal would have small but non-zero associated capacitance with it. Calculating such capacitances and either exploiting them or taking necessary measures to counteract it is a big deal in engineering electric circuits.

## Capacitors connected in Parallel and Series

Let's find the equivalent capacitance of capacitors in series and in parallel.

### Capacitors in parallel

Capacitance in two capacitors connected in parallel adds up, i.e.

#### Not-so-rigorous proof

When two capacitors are connected in parallel, then the terminals of capacitor will have the same voltage. So, if we swap the capacitors in parallel with some equivalent capacitor, it should have the same voltage drop as the either one of the parallel capacitors had. If we count the charges accumulated on the capacitors in parallel, they add up (If one capacitor had Q1 charges accumulated and the other Q2 then the equivalent charges accumulated is Q1+Q2). That means that the charges in the equivalent capacitor is the sum of charges accumulated... which means:
    $Q_{equivalent}=Q_{1}+Q_{2}...$
    $V_{equivalent}=V_{1}=V_{2}$ They are all equal, so let's call it "V".
       So,
    ${\frac {Q_{equivalent}}{V_{equivalent}}}={\frac {Q_{1}+Q_{2}}{V}}$
    ${\frac {Q_{equivalent}}{V_{equivalent}}}={\frac {Q_{1}}{V_{1}}}+{\frac {Q_{2}}{V_{2}}}$
       Therefore,
...We can generalize this for more than 2 capacitors - just keep adding the effective capacitance of each capacitor. So for five capacitors, the total capacitance would be
    $C_{equivalent}=C_{1}+C_{2}+C_{3}+C_{4}+C_{5}$

### Capacitors in series

Reciprocal of Capacitance adds up for capacitor connected in series., i.e.

#### (Once again) Not-So-Rigorous Proof

It's the exact opposite of parallel circuit.
First, the voltage drop must add up (for example, if two series capacitors C1 and C2 had voltage drop of 3V and 1V, then the equivalent capacitor had better have voltage drop of 4V).
What about the charge, however? The charges must remain the same in the equivalent capacitor. To illustrate, suppose two capacitors C1 and C2 are connected in series. Then if charge Q accumulates on one plate of C1, then charge of -Q would accumulate on the other plate. Conservation of charge dictates that the '-Q' must come from somewhere. That 'somewhere' is the top plate of C2.
So, the top plate of C2 loses '-Q' charge, which is essentially saying that C2 accumulates charge of 'Q'. Then, the other side of C2 will have a charge '-Q'. So, if we view the system holistically, the magnitude of charge accumulated on top of C1 is the magnitude of charge on bottom of C2)...
In short,
    $V_{equivalent}=V_{1}+V_{2}$
    $Q_{equivalent}=Q_{1}=Q_{2}$ They are all equal, so let's call it "Q".
       So,
    ${\frac {V_{equivalent}}{Q_{equivalent}}}={\frac {V_{1}+V_{2}}{Q}}$
    ${\frac {V_{equivalent}}{Q_{equivalent}}}={\frac {V_{1}}{Q_{1}}}+{\frac {V_{2}}{Q_{2}}}$
       Therefore,
    ${\frac {1}{C_{equivalent}}}={\frac {1}{C_{1}}}+{\frac {1}{C_{2}}}$
Once again, we can generalize this rule for more than 2 capacitors - just add the reciprocals!

## Capacitor as an energy storage element

A capacitor can be considered as a device that stores energy in the electric field by applying voltage across it. If we calculate the energy stored a capacitor (E) of capacitance C when voltage V is applied, we find that $E={\frac {CV^{2}}{2}}(J)$ (By the way, the magnetic analogue of this is called the inductor, and it possesses surprisingly similar characteristic with surprisingly similar equations.)

### Not-so-rigorous Proof

The power dissipated for an electric component was defined to be P=v i where v=voltage and i=current. Current is change of charge over time, or dQ/dt. We have defined C=Q/V, so Q=CV. Since C is constant, i = dQ/dt = C dV/dt. Plug this into the equation for power, and we get:
    $P=CV{\frac {dV}{dt}}$
Because power is rate at which energy is changing, (P=dW/dt), to find work W, we have to integrate with respect to time. This gives us:
    $W=\int Pdt=\int CV{\frac {dV}{dt}}dt$
Though mathematicians will be infuriated by what I'm about to say now, it usually works for most cases. If we consider derivatives like a fraction, then we note that the 'dt's will cancel out, giving us:
    $W=\int Pdt=\int CVdV$
       which gives us:
    $W={\frac {CV^{2}}{2}}$
...which is the work required to store charges in a capacitor with voltage V applied, which is the energy stored in the capacitor when we apply a voltage V.

## Capacitor with a Resistor (RC Circuits)

When we have a circuit with resistor and a capacitor, we have what is known as a RC circuit, which appears all the time in any electric system. It can be used to delay a signal or filter unwanted signals.

### Derivation

Let's consider a case where a resistor with resistance R is connected in series with a capacitor with capacitance C and a voltage source with voltage $-V_{source}$. Assume that the Capacitor at time=0 has potential difference $V_{0}$
If we take Kirchoff's Voltage Law for this circuit, what we will get is the following:
$-V_{source}+iR+V_{capacitor}=0$
We know that the current flowing through the resistor is same as the current flowing through the capacitor. Because Q=CV for capacitor, the current i is ${\frac {dQ}{dt}}=CV_{capacitor}'$. Replacing i, we get:
$-V_{source}+CV_{capacitor}'R+V_{capacitor}=0$
Add $V_{source}$ and divide by RC to get:
$V_{capacitor}'+{\frac {V_{capacitor}}{RC}}={\frac {V_{source}}{RC}}$
This is a first order differential equation. Solving this, we get:
$V_{capacitor}=Ae^{-{\frac {t}{RC}}}+B$
If we then differentiate this we get:
$V'_{capacitor}={\frac {-A}{RC}}e^{-{\frac {t}{RC}}}$
And limiting t towards infinity gives:
$V'_{capacitor}=0$
$V_{capacitor}=0+B$
If we plug this into the differential equation mentioned above, we will get:
$0+{\frac {B}{RC}}={\frac {V_{source}}{RC}}$
Thus, $B=V_{source}$
Now, plug in the equation that we've found for V(capacitor) for time=0 to find:
$V_{0}=A+V_{source}$
Which gives us $V_{0}-V_{source}=A$. Combined, we get:
We will call $\tau =RC$ which gives us:

### Interpretation of the Equation

At t=0, we can see that voltage of the capacitor is equal to its initial condition. We can also notice that as time approaches infinity, the exponential term gets smaller and smaller, which gives us voltage of the source. The nature of the function does not allow discontinuity, so that means that the function is slowly making a transition from V(0) to V(source). How fast? Just take the derivative.
With this interpretation, RC circuit is a 'circuit that makes a smooth transition from one voltage level to another in an exponential fashion.'

### Time Constant

$\tau$ is what is known as the "Time constant" of the RC circuit. It is a magnitude that indicates how slowly the circuit voltage is decreasing or increasing. Larger T implies longer transition between the two states.

### Practical uses for Time Constant

RC circuits are mainly used to create delays and filters.

#### Delay

Let's say you were making a switch where the user had to press a button for more than three seconds. Say this device was connected to some other machinery that considered anything higher than 4.5V as "ON." Also suppose you had a 5V voltage source. With just these information, you will be able to construct a RC circuit with appropriate time constant to achieve this effect. If we assume the capacitor is initially discharged (Vc(0)=0V), then it becomes a problem of mere algebraic manipulation.

#### Filter

Remember that high RC meant smoother transition. If the voltage source was changing (as in signals that comes in from a microphone), then what would happen?
Well, from waves we know that low sounds have low frequency. Low frequency means that it takes more time to change from one value to another. The opposite of that is high frequency, which changes its values rapidly. If our RC term is very high, then the RC circuit won't be able to "catch up" with the rapid transition of the high frequency. This means that the circuit will pass the low frequency signals better than the higher frequency ones.
Such use of RC circuit is called a Low-pass Filter and it has important applications in signal processing.