Charge and the coulomb

1. Electric charge is a property of matter, and in a metal circuit it is carried by the electrons that are free to move through the metal.
2. Charge is given the symbol $Q$ and its unit is the coulomb, written $C$.
3. One coulomb is a very large amount of charge next to the charge on a single electron, since about $6.25\times 10^{18}$ electrons must pass a point to deliver $1C$.
4. Charge is conserved, so none of it is created or destroyed on the way round a circuit and the charge that leaves the supply is the charge that comes back to it.
5. Charge is what carries energy from the supply to the components, which makes it the link between the energy transferred and the potential difference.
6. Small quantities of charge are often quoted in millicoulombs, where $1mC=1\times 10^{-3}C$.

Energy transferred by charge

1. When a charge $Q$ moves through a potential difference $V$, the energy transferred is given by $E=QV$.
2. In this equation $E$ is the energy transferred in joules $(J)$, $Q$ is the charge in coulombs $(C)$ and $V$ is the potential difference in volts $(V)$.
3. Applied to the cell, the equation counts the energy the supply gives to the charge, and $V$ is then the potential difference across the supply.
4. Applied to a component, the equation counts the energy the charge transfers as it passes through, and $V$ is then the potential difference across that one component.
5. The full chain has three links: the cell does work on the charge, the charge carries that energy round the circuit, and the charge transfers the energy inside the components.
6. Doubling the charge that passes doubles the energy transferred, and doubling the potential difference doubles it as well, because $E$ is proportional to each of them when the other is held fixed.
7. No time appears in $E=QV$, so the equation gives the energy a stated quantity of charge transfers and says nothing about how quickly that happens.

Using the equation

1. Begin every calculation by writing $E=QV$ down, before any numbers are substituted.
2. List the given data with their symbols and units, so that the missing quantity is obvious.
3. Substitute in joules, coulombs and volts, since the equation only balances in those units.
4. Multiply the charge by the potential difference, then attach the unit $J$ to the answer.
5. Quote the answer to the same number of significant figures as the least precise piece of data given.
6. Where the numbers are large, standard form keeps the working readable, so $1.2\times 10^{4}J$ is preferred to $12000J$.

Rearranging the equation

1. Rearranged for charge, the equation becomes $Q=\frac{E}{V}$.
2. Rearranged for potential difference, it becomes $V=\frac{E}{Q}$.
3. Read in words, $V=\frac{E}{Q}$ says that potential difference is the energy transferred per unit charge, which is exactly the definition of the quantity.
4. Putting the units into that form gives $1V=\frac{1J}{1C}=1{J\; C}^{-1}$, so a volt is a joule per coulomb.
5. Choose the version of the equation that already has the unknown on its own, which removes the need to rearrange after substituting.
6. A cell that transfers a known amount of energy to a known charge has a potential difference equal to the ratio of those two quantities, so an energy measurement can be used to find a potential difference.
7. Dividing a large energy by a small charge gives a large potential difference, which is a quick way of judging whether an answer is sensible.

Converting units first

1. The equation balances only when the energy is in joules, the charge is in coulombs and the potential difference is in volts.
2. Energy in kilojoules is multiplied by $1000$, so $2.4kJ=2400J$ and $0.75kJ=750J$.
3. Charge in millicoulombs is divided by $1000$, so $150mC=0.150C$.
4. Potential difference in millivolts is divided by $1000$, so $250mV=0.250V$.
5. Convert before substituting rather than afterwards, because mixing a prefix into the equation leaves the answer wrong by a factor of a thousand.

Multi-step problems

1. Some questions give a current and a time rather than a charge, so the charge has to be found from $Q=It$ before $E=QV$ can be used.
2. Keep the two stages separate on the page, writing each equation down above its own substitution.
3. Label each stage with the quantity it produces, so a charge in $C$ is never carried forward into a later line as though it were an energy in $J$.
4. In a series loop the same charge passes through every component, so once that charge is known the energy transferred in each component follows from $E=QV$ using that component's own potential difference.
5. Adding those separate energies gives the total energy the supply transferred to that charge, which is a useful check on the working.
6. Finish by testing the size of the answer, since a charge of a few coulombs moving through a few volts transfers tens of joules rather than thousands.