# OCR A-Level Physics/Fields, Particles and Frontiers of Physics/Capacitors

Capacitors are simply devices capable of storing electrical charge, and so capable of acting as a cell for a short period of time while discharging. They are made up of two metal plates, separated by an insulator, as illustrated by their circuit symbol (shown on the right).

## Capacitance

Capacitance has the symbol C and the unit for capacitance is Farad, F and it is usually expressed with prefixes because capacitance is usually very small. from this definition, the equation can be defined as $C={\frac {Q}{V}}$ However you will see it in the formula booklet as $Q=VC$

### Total Capacitance

The equations for finding the total capacitance are similar to the equations for total resistance.
For parallel we use $C_{total}=C_{1}+C_{2}+C_{3}+...$
For series we use ${\frac {1}{C_{total}}}={\frac {1}{C_{1}}}+{\frac {1}{C_{2}}}+{\frac {1}{C_{3}}}+...$

## Energy in Capacitors

=== P.d stored in a capacitor is proportional to the charge stored in it. If this was plotted, it would be a straight line through the origin. On a V against Q graph, the area under the graph is the energy stored by the capacitor. This is an area of a triangle where the base is charge and the height is p.d $W={\frac {1}{2}}QV$ From this we can derive another equation. $Q=VC$ === $W={\frac {1}{2}}(VC)V$
$W={\frac {1}{2}}CV^{2}$

## Charging and Discharging

When a capacitor charges or discharges, it does so exponentially. This means that when charging, it takes the same amount of time to double and when discharging, it takes the same amount of time to halve.

### Charging Capacitors

When a voltage is applied across a capacitor, electrons from one metal plate travel around the wire in the circuit to the other plate. By definition, the electrons can not travel across the insulator. For this reason, electrons are lost from one plate, which develops a positive charge, and added to another plate, which accumulates a negative charge.

### Discharging Capacitors

To discharge a capacitor, connect the charged capacitor to a resistor or an electrical component without a power supply. The current flows through the wire and the electrons reach equilibrium when the capacitor is fully discharged. The potential difference across the plates is zero. The time taken for the capacitor to discharge depends on:
* Capacitance of the capacitor.
* Resistance of the circuit.
Capacitance affects the amount of charge stored and the resistance affects the current flowing in the circuit. Due to capacitors discharging exponentially, we have the equation for voltage. $V=V_{0}e^{-{\frac {t}{CR}}}$
However the can be applied to charge also. $Q=Q_{0}e^{-{\frac {t}{CR}}}$ For this reason, the formulae booklet gives the formula as
$x=x_{0}e^{-{\frac {t}{CR}}}$

### Modelling decay

For a discharging capacitor, the p.d. across the capacitor and the p.d. across the resistor sum up to zero, according to Kirchhoff's Second Law. Therefore;
V$_R$=−V$_C${\textstyle V_{R}=-V_{C}}
Using Ohm's law and the capacitance equation, we derive the following;
IR=−Q/C {\textstyle IR=-{\frac {Q}{C}}}
Current is defined as the rate flow of charge therefore;
ΔQ/Δt R=−Q/C {\textstyle {\frac {\Delta Q}{\Delta t}}R=-{\frac {Q}{C}}}
Rearrangement and calculus;
ΔQ/Δt=−Q/RC {\textstyle {\frac {\Delta Q}{\Delta t}}=-{\frac {Q}{RC}}}
∫ 1/Q δQ=−∫ 1/RC δt{\textstyle \int {{\frac {1}{Q}}\delta Q}=-\int {{\frac {1}{RC}}\delta t}}
ln⁡Q=−t/RC+ln⁡k{\textstyle \ln Q=-{\frac {t}{RC}}+\ln k}
$Q=ke^{-{\frac {t}{CR}}}$
Arriving at the discharge equation;
Q=Q$_0$e$^{−t/CR}${\textstyle Q=Q_{0}e^{-{\frac {t}{CR}}}}

### Time Constant

Time constant, $\tau$, is the time it takes for a capacitor to discharge by 37% of Q$_0$. The symbol $\tau$, is the Greek letter 'tau'.
The time constant is also the time it takes a capacitor to charge by 63% of Q$_0$ The equation we are given for the time constant is $\tau =CR$ If we substituted
$\tau =CR$
Into
$x=x_{o}e^{-{\frac {t}{CR}}}$
Where
$t=\tau$ Then $x=x_{0}e^{-{\frac {CR}{CR}}}$ $x=x_{0}e^{-1}$
${\frac {x}{x_{0}}}={\frac {1}{e}}\approx 0.37$

### Parallel Plate Capacitor

The capacitance of a capacitor depends on both, the separation between the parallel plates and the area of overlap between the plates. Experimental results have shown that $C\propto {\frac {A}{d}}$. The constant of proportionality in this case is the permittivity of free space $\varepsilon _{0}$ which gives the equation:
$C={\frac {\varepsilon _{0}A}{d}}$
This is only true when the dielectric is a vacuum. Realistically, a vacuum will not be used so we use a relative permittivity for different dielectrics.
$\varepsilon =\varepsilon _{0}\varepsilon _{r}$
$\varepsilon _{r}$ depends on the dielectric. (You do not need to learn these values for the exam, they will be provided)
| Material | $\varepsilon _{r}$ |
| vacuum | 1 (by definition) |
| air | 1.0006 |
| perspex | 3.3 |
| mica | 4.0 |
| barium titanate | 1200 |
The general equation for a parallel plate capacitor is:
$C={\frac {\varepsilon A}{d}}$