Capacitance Flashcards
Mechanism of charging a capacitor
To move charge onto the plates of a capacitor, it must be connected to a voltage supply. The negative terminal of the supply pushes electrons onto one plate, making it negatively charged. Electrons are repelled from the other plate, making it positively charged. Figure 23.4 shows that there is a flow of electrons all the way round the circuit. The two ammeters will give identical readings. The current stops when the potential difference (p.d.) across the capacitor is equal to the electromotive force (e.m.f.) of the supply. We then say that the capacitor is ‘fully charged’.
Note: The convention is that current is the flow of positive charge. Here, it is free electrons that flow. Electrons are negatively charged; conventional current flows in the opposite direction to the electrons.
How to discharge a capacitor?
Discharging of a capacitor can be observed as follows. Connect the two leads of a capacitor to the terminals of a battery. Disconnect, and then reconnect the leads to a light-emitting diode (LED). It is best to have a protective resistor in series with the LED. The LED will glow briefly as the capacitor discharges.
How can charge stored on a capacitor be found?
The charge that flows past a point in a given time is equal to the area under a current-time graph (just as distance is equal to the area under a speed–time graph). So the magnitude of the charge on the plates in a capacitor is given by the area under the current–time graph recorded while the capacitor is being charged up.
Capacitance
C = Q / V
(Unit : Farad)
“The capacitance of a capacitor is the charge on the plates of the capacitor per unit potential difference across the plates”
To make the capacitor plates store more charge, we would have to use a supply of higher e.m.f. The greater the capacitance, the greater is the charge on the capacitor plates for a given potential difference across it.
Q = CV
This equation shows that the charge depends on two things: the capacitance C and the voltage V (double the voltage means double the charge). Note that it isn’t only capacitors that have capacitance. Any object can become charged by connecting it to a voltage. The object’s capacitance is then the ratio of the charge to the voltage.
Safety voltage marking on capacitors
Many capacitors are marked with their highest safe working voltage. If you exceed this value, charge may leak across between the plates, and the dielectric will cease to be an insulator. Some capacitors (electrolytic ones) must be connected correctly in a circuit. They have an indication to show which end must be connected to the positive of the supply. Failure to connect correctly will damage the capacitor, and can be extremely dangerous.
Energy stored in a capacitor
When we charge a capacitor, we use a power supply to push electrons onto one plate and off the other. The power supply does work on the electrons, so their potential energy increases. You recover this energy when you discharge the capacitor. In
order to charge a capacitor, work must be done to push electrons onto one plate and off the other. At first, there is only a small amount of negative charge on the left-hand plate. Adding more electrons is relatively easy, because there is not much repulsion. As the charge on the plate increases, the
repulsion between the electrons on the plate and the new electrons increases, and a greater amount of work must be done to increase the charge on the plate.
The graph in Figure 23.7a. shows how the p.d. V increases as the amount of charge Q increases. It is a straight line because Q and V are related by:
V = Q / C
The work done W in moving charge Q through a constant p.d. V. This is given by:
W = QV
Hence, the work done in charging a capacitor to a particular p.d. is given by:
W = 1/2 QV
Substituting Q = CV into this equation gives two further equations:
W = 1/2 CV^2
W = 1/2 Q^2 / C
(ref - Figure 23.7 : The area in a shows the energy stored in a capacitor; the area in b shows the energy required to drive a charge through a resistor)
Practical Activity 23.1 - Investigating energy stored in a capacitor
If you have a sensitive joulemeter (capable of measuring millijoules, mJ), you can investigate the equation for energy stored. A suitable circuit is shown in Figure 23.9.
The capacitor is charged up when the switch connects it to the power supply. When the switch is altered, the capacitor discharges through the joulemeter. (It is important to wait for the capacitor to discharge completely.) The joulemeter will measure the amount of energy released by the capacitor. By using capacitors with different values of C, and by changing the charging voltage V, you can investigate how the energy W stored depends on C and V.
Figure 23.9: With the switch to the left, the capacitor C charges up; to the right, it discharges through the joulemeter.
Capacitors in Parallel
When two capacitors are connected in parallel, their combined or total capacitance C total is simply the sum of their individual capacitances C1 and C2:
C total = C1 + C2
This is because, when two capacitors are connected together, they are equivalent to a single capacitor with larger plates. The bigger the plates, the more charge that can be stored for a given voltage, and hence the greater the capacitance.
Capacitors in parallel: deriving the formula
The total charge Q on two capacitors connected in parallel and charged to a potential difference V is simply given by:
Q = Ctotal × V
The total charge is given by the sum of these:
Q = Q1 + Q2 = C1V + C2V
Since V is a common factor:
Q = (C1 + C2)V
Comparing this with Q = Ctotal x V gives the required Ctotal = C1 + C2. It follows that for three or more capacitors connected in parallel, we have:
Ctotal = C1 + C2 + C3 + …
Capacitors in Series
1/Ctotal = 1/C1 + 1/C2
Here, it is the reciprocals of the capacitances that must be added to give the reciprocal of the total
capacitance. Notice that the total capacitance of two capacitors in series is less than either of the individual capacitances.
Capacitors in series: deriving the formula
This p.d. is divided (it is shared
between the two capacitors), so that the p.d. across C1 is V1 and the p.d. across C2 is V2. It follows that:
V = V1 + V2
Both capacitors are shown as storing the same charge Q. How does this come about When the voltage is first applied, charge −Q arrives on the left-hand plate of C1. This repels charge −Q off the right-hand plate, leaving it with charge +Q. Charge −Q now arrives on the left-hand plate of C2, and this in turn results in charge +Q on the right-hand plate.
Notice also that there is a central isolated section of the circuit between the two capacitors. Since this is initially uncharged, it must remain so at the end. This requirement is satisfied, because there is charge −Q at one end and +Q at the other (Figure 23.12). Hence, we conclude that capacitors connected in series store the same charge. This allows us to write equations for V1 and V2:
V2 = Q/C1 and V2 = Q/C2
The combination of capacitors stores charge Q when charged to p.d. V, and so we can write:
V = Q/Ctotal
Substituting these in V = V1 + V2 gives:
Q/Ctotal = Q/C1 + Q/C2
Cancelling the common factor of Q gives the required equation:
1/Ctotal = 1/C1 + 1/C2
If a capacitor is charged and then connected to a second capacitor, what happens to the charge and the energy that it stores?
When the capacitors are connected together, they are in parallel, because they have the same p.d. across them. Their combined capacitance Ctotal is equal to the sum of their individual capacitances.
Worked example 3 pg. no. 478 (must)
Capacitance of isolated spherical bodies
If we consider a conducting sphere of radius r insulated from its surroundings and carrying a charge Q it
will have a potential at its surface of V, where
V = 1/4πε0 × Q/r
Since , it follows that the capacitance of a sphere is C = 4πε0r.
Exponential decay
As charge flows off the capacitor, the potential difference reduces and so the current (the charge flowing per unit time) in the circuit also decreases. This type of decay is called exponential decay.
