7.2 - Capacitors

Question 1

What does an electric field do to charge

Answer 1

An electric field can cause charges to move. Indeed, this is why a current flows through a circuit - an electric field is set up within the conducting material and this causes electrons to feel a force and thus move through the wires and components of the circuit.

Question 2

What happens when there’s a gap in a circuit

Answer 2

Where there is a gap In a circuit, the effect of the electric field can be felt by charges across the empty space, but in general, conduction electrons are unable to escape their conductor and move across the empty space. This is why a complete conducting path is needed for a simple circuit to function.

Question 3

How could you test that an electric field will act across a space

Answer 3

An electric field will act across a space. You could text this by hanging a charged sphere near the plates and observing the fields force acting on the sphere.

Question 4

How can charge be made to flow in an incomplete circuit/ explain a capacitor

Answer 4

This can be demonstrated by connecting two large metal plates in a circuit with an air gap between them. When the power supply is connected, the electric field created In the conducting wires causes electrons to flow from the negative terminal towards the positive terminal. Since the electrons cannot cross the gap between the plates, they build up on the plate connected to the negative to the negative terminal, which becomes negatively charged. Electrons in the plate connected to the positive terminal flow towards the positive of the battery, resulting in positive charge being left on that plate. The attraction between the opposite charges across the gap creates an electric field between the plates, which increase until the potential difference across the gap is equal to the potential difference of the power SUPPLY.

A pair of plates such as this with an insulator between them is called a CAPACITOR

Question 5

What is a capacitor

Answer 5

A pair of plates with an insulator between them is called a capacitor. Charge will build up on a capacitor until the potential difference across the plates equals that provided by the power supply to which it is connected. At that stage it is said to be fully charged, and the capacitor is acting as a store of charge.

Question 6

What is capacitance

Answer 6

The amount of charge a capacitor can store, per volt applied across it, is called its capacitance, C, and is measured in farads (F)

Question 7

What’s the units of capacitance

Answer 7

Symbol C

Units farads (F)

Question 8

What does capacitance depend on

Answer 8

Capacitance depends on the size of the plates, their separation and the nature of the insulator between them.

Question 9

How can capacitance be calculated

Answer 9

Capacitance can be calculated by the equation

Capacitance (F) = charge stored (C) / potential difference across the capacitor (V)

C = Q/V

Question 10

How can we investigate stored charge

Answer 10

A device that will measure the amount of charge directly is called a coulomb-meter. By charging a capacitor to various different voltages, and discharging through the coulomb meter each time, you can verify the basic capacitor equation that C = Q/V

A graph of charge (on the y axis) against p.d (on the x axis) should produce a straight line through the origin. The gradient will equal the capacitance.

Question 11

What is a charged capacitor a store of

Answer 11

A charged capacitor is a store of electrical potential energy. When the capacitor is discharged, this energy can be transferred into other forms. Our definition of voltage gives the energy involved as E = QV.

The energy stored in a charged capacitor is given by E = 1/2 QV

^^ where has the missing half of the energy gone? This is a trick question, because our original equation assumes that the charge and voltage are constant. However, in order to charge a capacitor, it begins with zero charge stored on it and slowly fills up as the p.d increases, until the charge at voltage V is given by Q.

Question 12

How can we use a graph to calculate the energy stored on a charged capacitor

Answer 12

Each time we add a little extra charge, (triangle Q) this has to be done by increasing the voltage and pushing the charge on, which takes some energy (we are doing work)

By finding the area of each roughly rectangular strip between each addition of charge, we find V x Q, which is the amount of extra energy needed for that extra charge. Therefore, the sum of all the strips, or area under the graph/line, will give us the total energy stored. This is the area of a triangle, so it’s area is half base x height, so

E = 1/2 x Q x V

Question 13

What are the rearrangements of energy stored on a capacitor

Answer 13

E = 1/2 x Q x V

Because Q =CV, you can also find two other versions of this equation for the stored energy.

E =1/2 x (CV)V = 1/2CV^2

Or E = 1/2 x Q x (Q/C) = 1/2 x Q^2/C

Question 14

How can we investigate energy stored on a capacitor

Answer 14

You can investigate how the energy stored on a capacitor changes with the voltage used to charge it. Various combinations of identical series and parallel bulbs will have different overall resistances. If we add an extra parallel branch and increase the number of bulbs on each branch by one, we can keep the total resistance constant, but have more bulbs to light up.

By allowing our charged capacitor to discharge through these different groups of bulbs, and altering the voltage to keep the bulb brightness constant, we can confirm our equation
E = 1/2 x C x V^2 for energy stored on the capacitor

Question 15

Define capacitor

Answer 15

A capacitor is an electrical circuit component that stores charge and so can be used as an energy store

Question 16

Define capacitance

Answer 16

Capacitance is a measure of the capability of a capacitor, the amount of charge stored per unit voltage across the capacitor. It’s measured in farads, f

Question 17

How can we investigate current flow through a capacitor

Answer 17

You can investigate how the current through a capacitor changes over time by connecting a data logger, which senses current in series with the capacitor and then charge and discharge it. A suitable set up is shown on page 52 textbook, this set up could be altered to log the potential difference across the capacitor over time, using a voltage sensor in parallel across the capacitor. Make sure you have a good understanding of this practical as this method may be assessed in your examinations.

If the capacitor is fully charged, it’s voltage will be voltage of supply. We have already looked at how the electrons in the circuit are influenced by the electric field caused by a supply voltage, and their own mutual repulsion. When we discharge the capacitor, the rush of electrons is as high as it can be, the current starts at a maximum. After some electrons have discharged, the p.d across the capacitor is reduced and the electric field, and therefore the push on the remaining electrons, is weaker, current is less. Some time later the flow of electrons is so small that the current is down to trickle, eventually the capacitor will be fully discharged and there will be no more electrons moving from one side of capacitor to the other, current will be zero

Question 18

Tell me about the discharge curves for a capacitor

Answer 18

Three graphs, all the same curve/ trend

One with discharge current /A (y axis) and time on (x axis) it is downwards sloping, gradient is reducing over time

Another graph with p.d across capacitor/V (y axis) and time on x axis looks the same as the graph before

The last graph shown has charge on the capacitor/C (y axis) and time on the x axis, which again looks the same

^ all of these components decrease over time when discharging a capacitor

Question 19

How could we make a lamp light up for longer when we discharge the capacitor along this route, given the same power supply?

Answer 19

Store more charge on the capacitor

Decrease the rate at which the capacitor discharges

Eg. For the same max p.d, increasing the capacitance, C, will increase the charge stored as Q = CV. Alternatively, the charge would flow more slowly if the bulbs reisistance, R, was greater.

Question 20

How can an overall impression of the rate of discharge of a capacitor be gained

Answer 20

It can be gained by working out the time constant, fancy T symbol.

Question 21

How do we calculate time constant

Answer 21

T = RC, where resistance is In ohms and capacitance in farads, the answer is in seconds. The time constant tells you have many seconds it takes for the current to fall to 37% of its starting value. We will later see how the mathematics of 37% comes about. But for now we just need to understand that RC indicates how quickly a charged capacitor will discharge.

Question 22

Tell me about how capacitor discharge is used in car courtesy lights

Answer 22

Modern cars often have a light in the cabin that comes on when the door is opened, and remains on for a short time after the door is closed. This is useful in case it is dark, allowing the driver to see to put the key in the ignition. The light functions by having a capacitor discharge through the light bulb so that it dims and goes off as the charge runs out. In some cars, the length of time for which the light remains on after the door is closed is adjustable and can be set by the vehicle owner. (?? Who has these cars) this adjustable setting makes use of the idea of the time constant, RC. The owner will be able to adjust a switch connected to the courtesy light circuit, which connects more or less resistance to the discharging circuit. Thus for the same fully charged capacitor, the time taken to discharge completely will vary and the courtesy light illuminates the cabin for more or less time.

Question 23

What are the capacitor charging curves

Answer 23

By considering the charging process in the same way as we did the discharge of the capacitor, we can quickly work out that the charging process produces graphs such as these:

A graph of charging current/A (y axis) and time on x axis has a downwards sloping curve with gradient reducing as line approaches zero

A graph of p.d across the capacitor (y axis) and time on x axis was an upwards sloping curve and gradient reduced close to zero as p.d approached p.d of supply

A graph of charge on the capacitor (y axis) and time on x axis has an upwards sloping curve which flattens out as it reaches max charge

When charging a capacitor through a resistor, the time constant RC has exactly the same implications. A greater resistance or a larger capacitance, or both means the circuit will take longer to charge up the capacitor.

Question 24

Define the time constant

Answer 24

Time constant is, for a capacitor-resistor circuit, the product of the capacitance and the resistance, giving a measure of the rate for charging and discharging the capacitor. Symbol, tau (fancy T but sometimes just T instead)

Question 25

Describe what kind of graphs the charging and discharging of a capacitor graphs are

Answer 25

We have seen that the charging and discharging of a capacitor follows curving graphs in which the current is constantly changing, and so the rate of change of charge and p.d are also constantly changing. These graphs are known as exponential curves.

The shapes can be produced by plotting mathematical formulae which have power functions in them. In the case of discharging a capacitor, C, through a resistor, R, the function that describes the charge remaining on the capacitor, Q, at a time, t, is:

Q = Q(subscript zero) x e(^-t/RC)

Where Q(subscript 0) is the initial charge on the capacitor at t = 0, and e is the exponential function, which is used in the inverse function of natural logarithms e is roughly 2.718

Question 26

What’s the function that describes the charge remaining on the capacitor, Q, at a time t is:

Answer 26

Q = Q(subscript 0) x e(^-t/RC)

Where Q(subscript 0) = is the initial charge on the capacitor at time 0

Question 27

How can we rearrange the exponential decay equation

Q = Q(subscript 0) x e(^-t/RC) be rearranged

Answer 27

The p.d. across a discharging capacitor will fall as the charge stored falls. By substituting the equation Q = CV into our exponential decay equation, basically if we replace Q and Q(subscript 0) with CV we can cancel out C so the equation becomes

V = V(subscript 0) x e(^-t/RC)

Another rearrangement is..
Since V = IR,

We can input IR instead of V, and R cancels out so the equation becomes

I = I(subscript 0) x e(^-t/RC)

Question 28

How can we investigate using a spreadsheet to investigate the time constant

Answer 28

In order to create a timing circuit that fulfils the needs of a certain situation (such as a car courtesy light staying on for the desired length of time) we can model the circuit using a spreadsheet. This will then allow us to type in different possible values for the circuit components and see what the outcome will be, before building it.

You can create the spreadsheet without doing any experimentation. Give the cells the various formulae to calculate what capacitor theory tells us will happen, using the mathematics we have learnt. For example, the cell giving the time constant, T, does not require input from the user - it is programmed to display the multiplication of the capacitance and the discharge resistance. This value is then used in the formula for calculating the current column using equation I = I(subscript 0) x e(^-t/T) where T = time constant (RC)

You can then make a graph / discharge curve

Question 29

Tell me about using capacitor calculus

Answer 29

The mathematics used here, calculus (including integration) is beyond the scope of your examination specification in physics. However, for those studying high level mathematics, this will explain the source of the capacitor equation.

The equation for charge on a discharging capacitor is the solution to a differential equation based on considering the rules for voltages around the discharging circuit. With only the capacitor, C, and resistance, R, in the circuit, the emf is zero so:

0 = V(subscript C) + V(subscript R)

V(subscript C) = Q/C and V(subscript R) = IR

So

-IR = Q/C

Current is rate of change of charge so

I = triangle Q/ triangle T

So triangle Q/Q = -triangle t/RC

Integrating charge, Q with respect to Q(subscript 0) gives

ln Q = ln Q(subscript zero) - t/RC

Applying the inverse function of natural logarithms (e) gives

Q = Q(subscript 0) x e(^-t/RC)

Question 30

Tell me about the ‘37% life’

Answer 30

If we consider the charge at time T (time constant)

t = T = RC so -RC/RC = -1

So

Q = Q(subscript 0) x e(^-1)

e^-1 = 0.37

Q = Q(subscript 0) x 0.37

So the charge is 37% of its original value.

This shows that the time constant describes the decay of charge on a discharging capacitor in just the same way as radioactive half life describes the number of radioactive nuclei remaining , except that instead of describing the time taken to reach half of the initial value, T is the time taken to reach 37% of the initial value. This similarity comes from the fact that radioactive decay also follows an exponential equation N = N(subscript 0) x e(^-lander x t)

Question 31

Define exponential curves

Answer 31

Exponential curves are a mathematical function generated by each value being proportional to the value of one variable as the index of a fixed base: f(x) = b^x