Find source work or transferred charge from e.m.f.
Rearrange source e.m.f. = W/Q while keeping per-charge labels, total work and open-circuit cases distinct.
Section P4.2.3 · Electromotive force and potential difference
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Calculate source e.m.f. = W/Q.
- Rearrange a product with consistent units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What quantity results from V multiplied by C?
What you will learn
- Use W = e.m.f. × Q and Q = W/e.m.f.
- Interpret a source voltage without treating it as a fixed total energy.
Recover a total from the per-charge value
The source equation rearranges to W = e.m.f. × Q. A stated constant 9 V e.m.f. supplying work for 2 C gives 18 J. For a varying source account, one voltage reading does not automatically give the average work per charge for all the charge passed.
To find charge, Q = W/e.m.f. for a non-zero stated e.m.f. If a 4 V source does 12 J of electrical work, the matching passed charge is 3 C. Retain the source-work boundary rather than substituting only the useful output of one component.
Retain the model conditions
A voltage label is a per-charge value, so doubling Q at the same stated e.m.f. doubles W. It does not double e.m.f. A source with non-zero e.m.f. and zero passed charge can supply zero work during that open-circuit interval without losing its e.m.f.
If internal source losses or different time-dependent operating values are relevant, use the supplied complete source account and matching per-charge value. A numerical work result does not alone identify chemical capacity, useful energy efficiency or electrical power without additional information.
Sources for this lesson
Pause and explain
A 4 V source does 12 J of electrical work. What charge corresponds to this account?
Worked example
A source has constant e.m.f. 9 V and 2 C passes in its stated source account. Find the electrical work.
Show the worked solution
- Use W = e.m.f. × Q.
- Multiply 9 J/C by 2 C.
- W = 18 J.
Answer 18 J of source electrical work.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find source work for constant e.m.f. 9 V and passed charge 2 C.
Give me a hint
Multiply the source work per coulomb by the charge amount.
Compare my reasoning
- W = e.m.f. × Q.
- 9 × 2 = 18.
- The result is source work in joules.
18 J.
Look for these in your work
- Use the matching charge.
- State the source account.
Find source work at 6 V for 0.50 C and charge for 12 J at 4 V.
Give me a hint
Use a product for work and a quotient for charge.
Compare my reasoning
- W = 6 × 0.50 = 3 J.
- Q = 12/4 = 3 C.
- The results name different quantities.
3 J; 3 C.
Look for these in your work
- Select the right rearrangement.
- Keep work and charge units distinct.
A 9 V source has no charge pass during an open interval. Find the interval work under this model and assess a claim that its e.m.f. must therefore be zero.
Give me a hint
Use zero Q in the product without replacing a known source value.
Compare my reasoning
- W = 9 × 0 = 0 J.
- The source's stated e.m.f. is still 9 V.
- Zero transfer during the interval does not erase the source property.
0 J during the interval; zero e.m.f. does not follow.
Look for these in your work
- Separate interval work and source property.
- Avoid dividing an uninformative 0/0 account.
Common mistakes
- A per-charge value needs a matching charge amount.
- Use source work rather than one useful output.
- One instantaneous voltage need not represent a varying whole account.
What can you explain now?
At a fixed e.m.f., what happens to source work when the passed charge doubles?
Compare with the explanation
It doubles.
The fixed work-per-charge value multiplies the doubled charge amount.
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Make it stick
Find source work or charge and retain the distinction between an interval transfer and a source per-charge value.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.