E.m.f. and p.d — OCR A-Level Physics
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E.m.f. and p.d explained
Potential difference measures the energy transferred from electrical energy to other forms per unit charge as charge passes between two points in a circuit.
Read the full explanation
If a component transfers 12 J of energy when 4 C of charge passes through it, the p.d. across it is 12 J ÷ 4 C = 3 J C⁻¹, which is 3 V. The volt is therefore the joule per coulomb: 1 V = 1 J C⁻¹. In a series circuit the p.d. is shared between components; in a parallel circuit the p.d. across each branch is the same. A voltmeter is connected in parallel across the component being measured, because it must compare the energy at two points.
(b) electromotive force (e.m.f.) of a source such as a cell or a power supply
Electromotive force is the energy transferred from other forms to electrical energy per unit charge as charge passes through a source such as a cell or power supply. A cell that supplies 9 J of electrical energy for every 3 C of charge passing through it has an e.m.f. of 9 J ÷ 3 C = 3 J C⁻¹ = 3 V. The e.m.f. is a property of the source, not of the external circuit, and it is measured in volts for the same reason as p.d.: both are energy per unit charge. In a real cell, some of this energy is dissipated inside the source because of its internal resistance, so the terminal p.d. is less than the e.m.f. when current flows.
(c) distinction between e.m.f. and p.d. in terms of energy transfer
The distinction between e.m.f. and p.d. lies in the direction of energy transfer. E.m.f. is the energy transferred from other forms, such as chemical energy in a cell, to electrical energy per unit charge as charge passes through the source. P.d. is the energy transferred from electrical energy to other forms, such as thermal energy in a resistor, per unit charge as charge passes through a component. Both are measured in volts, so the unit alone cannot distinguish them; the definition must state which way the energy is transferred. In a circuit, the e.m.f. of the source supplies energy that is then transferred away by the p.d.s across the components.
(d) energy transfer; W = VQ ; W = EQ .
The equations W = VQ and W = EQ give the energy transferred when charge moves through a potential difference or an electromotive force. W is the energy transferred in joules, V is the p.d. in volts, E is the e.m.f. in volts and Q is the charge in coulombs. For example, if 5 C of charge passes through a p.d. of 12 V, the energy transferred is W = 12 V × 5 C = 60 J. Because 1 V = 1 J C⁻¹, the product of volts and coulombs is joules. The two equations have the same form because both e.m.f. and p.d. are energy per unit charge; the symbol used indicates whether the energy is supplied by a source or transferred by a component.
(e) energy transfer eV = 1/2 mv² for electrons and other charged particles.
When a charged particle is accelerated through a potential difference V, the electrical work done equals the kinetic energy gained. For a particle of charge q, the energy transferred is qV. In many problems the charge is the elementary charge e, so the energy transferred is eV. If the particle starts from rest, this energy becomes kinetic energy, so eV = ½mv², where m is the mass and v is the final speed. This equation links speed and accelerating voltage. For example, an electron accelerated through 100 V gains 100 eV, which is 1.60 × 10⁻¹⁷ J; using ½mv² gives a speed of about 5.9 × 10⁶ m s⁻¹. Remember that eV is an energy unit: 1 eV = 1.60 × 10⁻¹⁹ J.
Your focus
- Define potential difference as the energy transferred per unit charge between two points.
- State that the unit of p.d. is the volt and that 1 V = 1 J C⁻¹.
- Calculate a p.d. from energy transferred and charge using V = W ÷ Q, and describe how a voltmeter is connected to measure it.
Show all 15 objectives
- Define electromotive force as the energy transferred from other forms to electrical energy per unit charge in a source.
- State that e.m.f. is measured in volts and that 1 V = 1 J C⁻¹.
- Explain why the terminal p.d. of a real source is less than its e.m.f. when current flows.
- Distinguish between e.m.f. and p.d. by the direction of energy transfer per unit charge.
- Explain that both quantities are measured in volts despite representing opposite energy transfers.
- Apply the distinction to describe energy transfers in a simple circuit containing a source and components.
- Use W = VQ to calculate the energy transferred when charge passes through a potential difference.
- Use W = EQ to calculate the energy transferred when charge passes through an electromotive force.
- Check that quantities are in joules, volts and coulombs and that the correct symbol is used for p.d. or e.m.f.
- State that the energy transferred to a charged particle accelerated through a potential difference V is qV.
- Apply eV = ½mv² to find the speed, kinetic energy or accelerating voltage for electrons and other charged particles.
- Convert between electronvolts and joules using 1 eV = 1.60 × 10⁻¹⁹ J.
E.m.f. and p.d exam tips
Marking Points
- Potential difference is the energy transferred per unit charge between two points in a circuit.
- The unit of p.d. is the volt (V), where 1 V = 1 J C⁻¹.
- A p.d. of 3 V means 3 J of energy is transferred per coulomb of charge passing between the two points.
- A voltmeter is connected in parallel with the component so it measures the difference in electrical potential energy per unit charge across that component.
- In a series circuit the total p.d. is shared between components; in a parallel circuit the p.d. across each branch is equal.
- Electromotive force is the energy transferred from other forms to electrical energy per unit charge passing through a source.
- The e.m.f. of a source such as a cell or power supply is measured in volts, where 1 V = 1 J C⁻¹.
- The e.m.f. is a property of the source itself and represents the energy supplied per unit charge.
- For a real source, the terminal p.d. is less than the e.m.f. when current flows because energy is dissipated inside the source.
- The e.m.f. can be measured by connecting a high-resistance voltmeter across the source when no current flows.
- E.m.f. is the energy transferred from other forms to electrical energy per unit charge in a source.
- P.d. is the energy transferred from electrical energy to other forms per unit charge across a component.
- Both e.m.f. and p.d. are measured in volts, so the distinction is the direction of energy transfer, not the unit.
- In a circuit, the total energy supplied per unit charge by the source is transferred away by the components.
- The e.m.f. is associated with a source, whereas p.d. is measured between two points across a component.
- W = VQ gives the energy transferred when charge Q passes through a potential difference V.
- W = EQ gives the energy transferred when charge Q passes through an electromotive force E.
- In both equations W is in joules, Q is in coulombs, and V or E is in volts.
- The equations work because 1 V = 1 J C⁻¹, so multiplying volts by coulombs gives joules.
- The same relationship applies to e.m.f. and p.d. because both are energy per unit charge; the symbol used shows whether energy is supplied or transferred away.
- Electrical energy transferred to a charged particle accelerated through potential difference V is qV, or eV when the charge magnitude is the elementary charge e.
- For a particle starting from rest, the energy transferred equals the kinetic energy gained, so eV = ½mv².
- The electronvolt is an energy unit: 1 eV = 1.60 × 10⁻¹⁹ J.
- Use the correct mass and charge for the particle: for an electron, m = 9.11 × 10⁻³¹ kg and e = 1.60 × 10⁻¹⁹ C.
- When using eV = ½mv² with mass in kg and speed in m s⁻¹, you must convert the energy in eV to joules before calculating.
Examiner Tips
- 💡Write the unit as V and, when asked to show it is derived, substitute 1 V = 1 J C⁻¹ and expand the joule into base units.
- 💡Use W = VQ to check numerical answers: rearrange to V = W ÷ Q and confirm the unit is J C⁻¹.
- 💡Read the circuit diagram carefully to decide whether the p.d. is across a single component or shared between several in series.
- 💡Use the definition 'energy transferred per unit charge' and state the unit as V, showing 1 V = 1 J C⁻¹ when required.
- 💡When comparing e.m.f. and terminal p.d., mention internal resistance and the energy dissipated inside the source.
- 💡Remember that a high-resistance voltmeter across a source on open circuit reads approximately the e.m.f.
- 💡In comparison questions, state the direction of energy transfer for each quantity rather than only giving the unit.
- 💡Use a concrete example, such as a cell charging a resistor, to show e.m.f. supplying energy and p.d. transferring it away.
- 💡Check that your answer mentions both quantities and the per-unit-charge basis of each.
- 💡Write down the equation, substitute values with units, and state the answer in joules.
- 💡If charge is given in millicoulombs or microcoulombs, convert to coulombs before multiplying.
- 💡Use W = VQ for energy transferred by a component and W = EQ for energy supplied by a source, and label the symbol accordingly.
- 💡Write the energy equation first, then substitute values with units; this makes the conversion from eV to J clear.
- 💡Check whether the question asks for speed, kinetic energy or accelerating voltage, and rearrange before calculating.
- 💡Keep at least three significant figures in intermediate values and round only the final answer.
- 💡If the particle is not an electron, look up or use the given charge and mass rather than assuming electron values.
Common Mistakes
- Confusing p.d. with current: current is the rate of flow of charge measured in amperes, whereas p.d. is energy transferred per unit charge measured in volts.
- Connecting a voltmeter in series: a voltmeter must be in parallel with the component, because it compares the energy per unit charge at two points.
- Thinking the volt is a base SI unit: the volt is a derived unit equal to 1 J C⁻¹, equivalent to 1 kg m² s⁻³ A⁻¹.
- Thinking e.m.f. is a force: despite the name, e.m.f. is an energy per unit charge measured in volts, not a force in newtons.
- Assuming the terminal p.d. always equals the e.m.f.: when current flows, energy is dissipated inside the source, so the terminal p.d. is lower.
- Treating e.m.f. as a property of the external components: e.m.f. belongs to the source, whereas p.d. is measured across components in the circuit.
- Saying e.m.f. and p.d. differ because they have different units: both are measured in volts, so the difference is the direction of energy transfer.
- Describing p.d. as energy supplied to the circuit: p.d. is energy transferred from electrical energy to other forms, whereas e.m.f. is energy transferred to electrical energy.
- Using the terms interchangeably in circuit calculations: keep e.m.f. for the source and p.d. for components to make energy transfers clear.
- Using the wrong unit for charge: Q must be in coulombs, so convert millicoulombs or microcoulombs before substituting.
- Mixing up V and E in the two forms: use V for a p.d. across a component and E for the e.m.f. of a source.
- Forgetting that the product VQ has units of joules: check by substituting 1 V = 1 J C⁻¹ so that V × C = J.
- Using the electron mass for a proton or other charged particle. Correction: use the mass of the specific particle named in the question.
- Forgetting to convert electronvolts to joules before substituting into ½mv². Correction: multiply eV by 1.60 × 10⁻¹⁹ J eV⁻¹ to obtain energy in joules.
- Assuming the particle starts from rest when it may have an initial speed. Correction: check the wording; if it does not start from rest, include the initial kinetic energy.
- Confusing potential difference V with speed v. Correction: keep the symbols distinct and check which quantity the question gives.