Syllabus coverage
- State that there are positive and negative charges, that positive charges repel other positive charges, negative charges repel other negative charges, but positive charges attract negative charges
- Describe simple experiments to show the production of electrostatic charges by friction and to show the detection of electrostatic charges
- Explain that charging of solids by friction involves only a transfer of negative charge (electrons), describe an experiment to distinguish between electrical conductors and insulators, and recall and use a simple electron model to explain the difference, with typical examples
- State that charge is measured in coulombs, describe an electric field as a region in which an electric charge experiences a force, and state that the direction of an electric field at a point is the direction of the force on a positive charge at that point
- Describe simple electric field patterns, including the direction of the field, around a point charge, around a charged conducting sphere, and between two oppositely charged parallel conducting plates
- Know that electric current is related to the flow of charge, describe the use of ammeters (analogue and digital) with different ranges, describe electrical conduction in metals in terms of the movement of free electrons, and know the difference between direct current (d.c.) and alternating current (a.c.)
- Define electric current as the charge passing a point per unit time, recall and use the equation I = Q / t, state that conventional current is from positive to negative and that the flow of free electrons is from negative to positive
- Define electromotive force (e.m.f.) as the electrical work done by a source in moving a unit charge around a complete circuit, know that e.m.f. is measured in volts (V), and recall and use the equation E = W / Q
- Define potential difference (p.d.) as the work done by a unit charge passing through a component, know that the p.d. between two points is measured in volts (V), describe the use of voltmeters (analogue and digital) with different ranges, and recall and use the equation V = W / Q
- Recall and use the equation for resistance R = V / I, describe an experiment to determine resistance using a voltmeter and an ammeter and do the appropriate calculations, and state, qualitatively, the relationship of the resistance of a metallic wire to its length and to its cross-sectional area
- Sketch and explain the current-voltage graphs for a resistor of constant resistance, a filament lamp and a diode, and recall and use the relationships that the resistance of a metallic conductor is directly proportional to its length and inversely proportional to its cross-sectional area
- Understand that electric circuits transfer energy from a source of electrical energy, such as an electrical cell or mains supply, to the circuit components and then into the surroundings; recall and use the equations for electrical power P = IV and for electrical energy E = IVt; define the kilowatt-hour (kW h) and calculate the cost of using electrical appliances where the energy unit is the kW h
Part of the IGCSE Physics syllabus 0625.
Electric charge: attraction and repulsion
There are two kinds of electric charge, positive and negative, and the force between them follows one simple rule. Like charges repel and unlike charges attract: two positive charges push each other away, two negative charges push each other away, but a positive and a negative charge pull towards each other. Charge is measured in coulombs (C), and the force between two charges becomes weaker as the distance between them increases.
A simple experiment shows the rule. Rub a polythene rod with a woollen cloth and hang it from a thread, then bring a second rubbed polythene rod close to it. The rods swing apart, because both now carry the same charge and repel each other. Bring up a perspex rod that has been rubbed instead, and the suspended rod swings towards it, because the two rods carry opposite charges and attract each other.
The charges are produced by friction. Rubbing two insulators together transfers electrons from one surface to the other. The material that gains electrons becomes negatively charged, and the material that loses them is left with more protons than electrons, so it becomes positively charged. Only the electrons move: positive charge is fixed in the nuclei of the atoms and is never transferred when two solids are rubbed together.
To detect a charge you use a gold-leaf electroscope. A charged rod is touched on, or held near, the metal cap. The charge spreads over the metal stem and the gold leaf, so both carry the same charge, the leaf is repelled and it rises. The larger the charge, the further the leaf rises. A charged rod also attracts small uncharged objects such as scraps of paper, because it pulls the opposite charge inside them towards the near surface.
- Like charges repel and unlike charges attract.
- Charge is measured in coulombs (C).
- Rubbing insulators transfers electrons only: gain electrons and the material becomes negative, lose them and it becomes positive.
- A gold-leaf electroscope detects charge, and the leaf rises because like charges repel.
Conductors and insulators: the simple electron model
Whether a material can carry charge depends on how tightly its electrons are held. In a metal, the outer electrons of each atom are not tied to any one atom. They form a sea of free (delocalised) electrons that can move through the structure, so metals are electrical conductors and charge can flow through them.
In an insulator, every electron is held firmly by its own atom, so there are no free electrons and charge cannot flow. This is why a plastic covering keeps a cable safe to touch, and why a charged polythene rod keeps its charge instead of losing it the moment you pick it up with your hand.
You can test a material with a simple circuit: a cell, a lamp and a gap for the test material. If the material is a conductor the lamp lights, because free electrons carry charge round the circuit. If it is an insulator the lamp stays dark. A more sensitive test uses a charged electroscope: touch the metal cap with the material. A conductor lets the charge flow away to earth through your body, so the leaf falls; an insulator traps the charge on the cap, so the leaf stays up.
Typical conductors are copper, aluminium, silver, brass, mercury, salty water and graphite, which is an unusual non-metal that still conducts. Typical insulators are polythene, rubber, glass, dry wood, dry air and most plastics. The difference is not that a conductor contains charge and an insulator does not - both are full of electrons - it is whether those electrons are free to move.
- Conductor: has free electrons that can move, so charge flows through it.
- Insulator: every electron is held by its own atom, so charge cannot flow.
- Conductors: copper, aluminium, silver, brass, mercury, salty water, graphite.
- Insulators: polythene, rubber, glass, dry wood, dry air, most plastics.
Electric fields
Charges attract and repel each other without touching, so the force must be carried across the space between them. That region is the electric field: a region in which an electric charge experiences a force.
Field lines are drawn to show a field. They point in the direction of the force on a positive charge, so field lines always run away from a positive charge and towards a negative charge. The lines are drawn closer together where the field is stronger, and they never cross one another, because at any point there is only one direction in which the force can act.
Around a point charge, the field lines are straight and radiate outwards in every direction, with the arrowheads pointing away from a positive charge. Around a single negative point charge the lines point inwards instead. A charged conducting sphere looks exactly the same from the outside: the lines are radial and they leave or enter the surface at right angles to it.
Between two oppositely charged parallel plates, the field is quite different. The lines are straight, parallel and evenly spaced, and they run from the positive plate to the negative plate, so the field is uniform in the gap between them. The force on a charge is the same everywhere in that gap, which is why parallel plates are used when a steady force on a charge is wanted. The field near the edges is not uniform, but end effects are not examined.
- An electric field is a region in which a charge experiences a force.
- Field lines point in the direction of the force on a positive charge: out of a positive charge, into a negative charge.
- Lines closer together mean a stronger field, and field lines never cross.
- Between oppositely charged parallel plates the field is uniform.
Electric current: charge on the move
An electric current is a flow of charge. In a metal the charge that moves is the free electrons, which drift through the metal. Current is measured with an ammeter, which is connected in series with the component so that the current passes through it. Analogue and digital ammeters both come with several ranges, such as 0-1 A and 0-5 A, and you should choose the range that puts the reading near the middle of the scale, because that gives the most accurate value.
Current is defined as the charge passing a point per unit time, which gives the equation I = Q / t. Here I is the current in amperes (A), Q is the charge in coulombs (C) and t is the time in seconds (s). One ampere is one coulomb per second. Rearranged, Q = I x t and t = Q / I.
Worked example. A current of 0.50 A flows through a lamp for 3.0 minutes. Convert the time to seconds first: t = 3.0 x 60 = 180 s. Then Q = I x t = 0.50 x 180 = 90 C. Always convert minutes into seconds before you substitute, because the equation only works with seconds.
There are two kinds of current. Direct current (d.c.) always flows in the same direction, and it is produced by cells and batteries. Alternating current (a.c.) repeatedly reverses its direction, and it is the form supplied by the mains, changing direction 50 times every second (50 Hz). Whichever kind is flowing, conventional current is taken to flow from the positive terminal to the negative terminal, while the free electrons actually flow from the negative terminal to the positive terminal. The two directions are opposite, but every equation and every ammeter reading uses conventional current.
- Current is the charge passing a point per unit time: I = Q / t, with Q in coulombs, t in seconds and I in amperes.
- Ammeters are connected in series and have several ranges; choose the range that gives a reading near the middle of the scale.
- d.c. flows in one direction and is produced by cells; a.c. reverses direction and is supplied by the mains at 50 Hz.
- Conventional current flows from positive to negative; the free electrons flow from negative to positive.
Electromotive force and potential difference
A cell does work on the charge that it pushes round a circuit, and the energy comes from its chemical store. The electromotive force (e.m.f.) of a source is the electrical work done by the source in moving a unit charge around a complete circuit. It is measured in volts (V). Despite the name, e.m.f. is not a force: it is the energy transferred to each coulomb of charge, and one volt is one joule per coulomb.
Potential difference (p.d.) is the same idea applied to a component. It is the work done by a unit charge passing through the component, and it is also measured in volts (V), this time with a voltmeter connected in parallel across the component. A p.d. of 3 V across a lamp means that 3 J of energy is transferred out of the electrical store for every coulomb of charge that passes through the lamp.
The two equations are E = W / Q for a source and V = W / Q for a component, where W is the work done, or energy transferred, in joules, Q is the charge in coulombs, and E and V are in volts. Because the volt is a joule per coulomb, the unit can also be written 1 V = 1 J/C.
Worked example. A cell transfers 36 J of energy to the charge that passes through it, and 6.0 C flows round the circuit. Then E = W / Q = 36 / 6.0 = 6.0 V. If the same 6.0 C then passes through a lamp and transfers 12 J to it, the p.d. across the lamp is V = W / Q = 12 / 6.0 = 2.0 V. In a series circuit the e.m.f. of the source is shared between the components, so the voltages round the loop always add up.
- e.m.f.: the electrical work done by a source in moving a unit charge around a complete circuit; measured in volts (V).
- p.d.: the work done by a unit charge passing through a component; measured in volts (V).
- E = W / Q and V = W / Q, with W in joules and Q in coulombs. One volt is one joule per coulomb.
- Voltmeters are connected in parallel across a component; ammeters are connected in series.
Resistance, how to measure it, and the current-voltage graphs
Resistance is a measure of how hard it is to push a current through a component. It is defined by the equation R = V / I, where R is the resistance in ohms, V is the potential difference across the component in volts and I is the current through it in amperes. A component with a high resistance needs a large p.d. to drive even a small current. Rearranged, V = I x R and I = V / R.
Experiment to find a resistance. Connect the component in series with a cell, a variable resistor and an ammeter, and connect a voltmeter in parallel across the component. Adjust the variable resistor to give a series of readings, and for each one record the current I and the p.d. V. Then calculate R = V / I for each pair of readings and take an average. Choose meter ranges that put the needle near the middle of the scale, because a reading taken near one end of a range is much less accurate.
How a wire's resistance depends on the wire. For a metallic conductor the resistance is directly proportional to its length: double the length and the resistance doubles, because the electrons have to push past twice as many ions. The resistance is inversely proportional to the cross-sectional area: double the area and the resistance halves, because there is a wider path for the electrons. A long, thin wire therefore has a much higher resistance than a short, thick wire of the same material.
Current-voltage graphs. Plotting current on the vertical axis against voltage on the horizontal axis shows how a component behaves. A fixed resistor at constant temperature gives a straight line through the origin, so the current is proportional to the voltage and the resistance is constant. A filament lamp gives a curve that flattens as the voltage rises: the current heats the filament, the metal ions vibrate more vigorously and the electrons collide with them more often, so the resistance increases and the same increase in voltage produces a smaller increase in current. A diode conducts in one direction only: in the forward direction almost no current flows until the voltage reaches about 0.6 V, after which the current rises steeply, while in the reverse direction its resistance is so high that the current is negligible.
- R = V / I, with R in ohms, V in volts and I in amperes.
- Resistance is directly proportional to length and inversely proportional to cross-sectional area.
- Fixed resistor: straight line through the origin, so the resistance is constant.
- Filament lamp: the resistance rises as the filament heats, so the graph curves.
- Diode: conducts in one direction only, above about 0.6 V.
| Component | Shape of the current-voltage graph | What the shape tells you |
|---|---|---|
| Fixed resistor at constant temperature | A straight line through the origin | The current is proportional to the voltage, so the resistance is constant |
| Filament lamp | A curve that flattens as the voltage rises | The resistance increases as the filament gets hotter |
| Diode | No current in the reverse direction; a steep rise after about 0.6 V in the forward direction | The diode has a very high resistance in reverse and conducts in one direction only |
Electrical power and electrical energy: P = IV and the kilowatt-hour
An electric circuit is a way of transferring energy. Energy is transferred from a source, such as an electrical cell or the mains supply, to the circuit components and then out into the surroundings. A lamp transfers energy from the electrical store to the internal (thermal) store and to light, and a motor transfers it to the kinetic store. The rate at which a component transfers energy is its power.
The equation for electrical power is P = IV, where P is the power in watts (W), I is the current in amperes and V is the potential difference in volts. Because power is energy per second, the electrical energy transferred is E = IVt, where t is the time in seconds and E is in joules. A kettle connected to 230 V that draws 9.0 A has a power of P = IV = 9.0 x 230 = 2070 W, which is about 2.1 kW.
For household energy a joule is inconveniently small, so we use the kilowatt-hour (kW h), which is the energy transferred by a device of power 1 kilowatt in 1 hour. Worked example: a 60 W lamp is left on for 5.0 hours. In kilowatts the power is 60 / 1000 = 0.060 kW, so E = 0.060 x 5.0 = 0.30 kW h. One kilowatt-hour is 3600000 J, or 3.6 MJ, because 1000 W x 3600 s = 3600000 J.
Cost. The cost of using an appliance is cost = energy in kW h x cost per kW h. At 20 p per kW h, leaving that 60 W lamp on for 5.0 hours costs 0.30 x 20 = 6.0 p. To use this method you must have the power in kilowatts (divide the watts by 1000) and the time in hours (divide the minutes by 60), because the kilowatt-hour is built from kilowatts and hours, not watts and seconds.
- P = IV, with P in watts, I in amperes and V in volts.
- E = IVt, with E in joules and t in seconds.
- One kilowatt-hour is the energy transferred by a 1 kW device in 1 hour: 1 kW h = 3.6 MJ.
- cost = energy in kW h x cost per kW h; convert watts to kilowatts and minutes to hours first.
Key terms you must be able to define
- Electric charge
- A property of matter that comes in two kinds, positive and negative; like charges repel and unlike charges attract. Charge is measured in coulombs (C).
- Electric field
- A region in which an electric charge experiences a force; its direction at a point is the direction of the force on a positive charge placed at that point.
- Conductor
- A material that contains free electrons able to move through it, so charge can flow; examples are copper, aluminium, silver, brass and graphite.
- Electric current
- The charge passing a point per unit time: I = Q / t. It is measured in amperes (A), where 1 A = 1 C/s.
- Electromotive force (e.m.f.)
- The electrical work done by a source in moving a unit charge around a complete circuit: E = W / Q. It is measured in volts (V), and 1 V = 1 J/C.
- Potential difference (p.d.)
- The work done by a unit charge passing through a component: V = W / Q. It is measured in volts (V), and 1 V = 1 J/C.
- Resistance
- A measure of how hard it is to push a current through a component: R = V / I. It is measured in ohms, and for a metallic conductor it is directly proportional to length and inversely proportional to cross-sectional area.
- Kilowatt-hour (kW h)
- The energy transferred by a device of power 1 kilowatt in 1 hour; 1 kW h = 3600000 J = 3.6 MJ. It is the unit used to calculate the cost of electricity.
Applied to the real world
Cambridge sets physics in unfamiliar contexts - a car braking, a power station, a hospital scanner, a satellite. Use these as models for how to apply the physics, then write your own.
A car battery has an e.m.f. of 12 V, which means it transfers 12 J of energy to every coulomb of charge it pushes round the circuit. A headlamp rated at 60 W therefore draws a current of I = P / V = 60 / 12 = 5.0 A, and its resistance is R = V / I = 12 / 5.0 = 2.4 ohms. The battery's chemical store empties as it does this electrical work, and the lamp transfers the energy to its surroundings by heating and by light.
A kettle is connected to the 230 V mains and draws a current of 9.0 A, so its power is P = IV = 9.0 x 230 = 2070 W, about 2.1 kW. Its resistance is R = V / I = 230 / 9.0 = 26 ohms to 2 significant figures. The heating element is a long, thin wire of high resistance, which is why it glows red hot while the thick copper cable feeding it stays cool.
A 2.0 kW electric heater runs for 6.0 hours a day. In one day it transfers E = 2.0 kW x 6.0 h = 12 kW h, which is 12 x 3.6 MJ = 43 MJ. Multiplying the 12 kW h by the cost of one kilowatt-hour gives the daily running cost, and the same calculation shows how much energy and money is saved by turning the thermostat down or switching the heater off an hour earlier.
Friction between ice particles swirling inside a storm cloud transfers electrons, so the bottom of the cloud becomes negatively charged while the ground below it is left positively charged. The electric field between cloud and ground grows until the air, which is an insulator, can no longer hold the charges apart. The charge then flows in a lightning strike: a current of tens of thousands of amperes lasting a fraction of a second.
Petrol flowing through a hose rubs against the inside of the pipe and becomes electrostatically charged, exactly as a polythene rod does when it is rubbed with a cloth. A spark from that charge could ignite the vapour, so fuel tankers and aircraft being refuelled are bonded and earthed with a metal strap. The strap is a conductor, so the charge flows harmlessly to the ground instead of building up.
Flashcards
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Exam-style questions
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Exam technique for this topic
- Give every answer its unit: coulombs for charge, amperes for current, volts for e.m.f. and p.d., ohms for resistance, watts for power, joules for energy and kW h for the energy on a bill. A bare number usually loses the mark.
- Convert the time to seconds before using I = Q / t or E = IVt. For a cost calculation the time must be in hours instead, and the power in kilowatts, because the kilowatt-hour is built from kilowatts and hours.
- Connect the ammeter in series and the voltmeter in parallel across the component. In a resistance experiment the voltmeter goes across the component under test, not across the ammeter or the whole circuit.
- In a current-voltage question, say what the shape means for the resistance. The filament lamp curves because its resistance increases as the filament gets hotter, and a diode conducts in one direction only.
- Remember both directions: conventional current flows from positive to negative, while the free electrons flow from negative to positive. If the question asks about electron flow, that second statement is the one being tested.
- In R = V / I use the p.d. across that component and the current through it. Mixing a voltage from one part of a circuit with a current from another part is the commonest error in these calculations.
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Where this fits in IGCSE Physics
This page is one topic of Cambridge IGCSE Physics 0625. IGCSE Physics is the course home: every area and every topic in Cambridge's own order, the papers you sit and what each one tests, and the exam-practice tools. The full topic list, with each topic linked to its revision page, is on the IGCSE Physics syllabus page.