Please read the extension material for a more in depth treatment of the topic
Learning Objective 

Capacitance is defined as:
The charge required to cause unit potential difference in a conductor.
Capacitance is measured in units called farads (F) of which the definition is:
1 Farad is the capacitance of a conductor, which has potential difference of 1 volt when it carries a charge of 1 coulomb.
So we can write from this definition:
Capacitance (F) = Charge (C)
Voltage (V)
In code, this is written:
[Q  charge in coulombs (C); C – capacitance in farads (F); V  potential difference in volts (V)]
We can show the relationship between the voltage and the charge on the graph.
The charge is directly proportional to the current. This means that it's a straight line with a positive gradient, going through the origin. This stands to reason  no voltage, no charge. Capacitance is the gradient of the graph.
Explain how the graph shows that voltage and charge are directly proportional. 
The voltage rises as we charge up a capacitor, and falls as the capacitor discharges. The current falls from a high value as the capacitor charges up, and falls as it discharges.
At its simplest the RC network is a series circuit consisting of a capacitor and a resistor connected to a source.
If we discharge a capacitor, we find that the charge decreases by the same fraction for each time interval. If it takes time t seconds for the charge to decay to 50 % of its original level, we find that the charge after another t seconds is 25 % of the original (50 % of 50 %). This time interval is called the halflife of the decay. The decay curve against time is called an exponential decay.
The voltage, current, and charge all decay exponentially during the capacitor discharge.
We can note the voltage and current at time intervals and plot the data, which gives us the exponential graph, using a circuit like this.
The graph is like this:
We should note the following about the graph:
Its shape is unaffected by the voltage.
The halflife of the decay is independent of the voltage.
The product RC (capacitance × resistance) is called the time constant. The units for the time constant are seconds. We can go back to base units to show that ohms × farads are seconds.
After RC seconds the voltage is 37 % of the original. To increase the time taken for a discharge we can:
Increase the resistance.
Increase the capacitance.
The halflife is 69 % of the time constant.
t_{1/2} = 0.69 RC
Electronic engineers use the time constant in preference to the halflife. In theory the exponential decay should never allow a capacitor to discharge completely, but in practice, a rule of thumb is that the capacitor is discharged completely after 5 RC seconds. In some text books, the physics code for time constant is t ('tau' a Greek lower case letter 't').
If you want to find the derivations for these equations, please follow the extension link.
Worked example A 5000 mF capacitor is charged to 12.0 V and discharged through a 2000 W resistor. (a) What is the time constant? (b) What is the halflife of the decay? 
Answer (a) Time constant = RC = 2000 W × 5000 × 10^{6} F = 10 s. (b) t_{1/2 } = 0.693 × RC = 0.693 × 10 = 6.93 s. 
After 50 seconds, this capacitor will have been discharged completely.
A capacitor of 470 mF is discharged through a 150 k resistor, with a starting voltage of 12 V? (a) What is the time constant? (b) What is the time needed to get to 6 V? (c) What is the time needed to get to 3 V? (d) What is the voltage across the capacitor after RC seconds?

When we charge up a capacitor, we get an exponential rise in charge and voltage. We get an exponential fall in the current. This is because when we start to charge up the capacitor, the current is a maximum and the voltage is zero. When the voltage is at a maximum, the current is zero, because no charge can flow on.
The graphs are like this:
After RC seconds the capacitor has charged up to 63 % of its final voltage.
As in a discharge there is a halflife in the charging of a capacitor; we can relate it to the time constant by the relationship:
t_{1/2} = 0.69 RC
After 5 RC time constants, the capacitor is almost completely charged up, so the voltage is almost V_{0}. In practice, the voltage is close to V_{0} before this.
A capacitor of 470 mF is charged up through a 150 k resistor, with a final voltage of 12 V? (a) What is the time constant? (b) What is the time needed to get to 6 V? (c) What is the voltage across the capacitor after RC second. (d) How long does it take for the capacitor completely to charge? 