Use coulombs and signed charge prefixes
State the charge unit and convert supplied micro-, nano- and millicoulomb values while preserving their signs.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Follow the selected magnetism or charge section. Field lines are explanatory models, not material threads or measured force values. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Use powers of ten and signed sums.
- Distinguish electric charge from other physical quantities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which is the larger positive magnitude: 10⁻³ or 10⁻⁶?
What you will learn
- State that charge is measured in coulombs (C).
- Convert supplied signed prefix values to consistent coulomb units.
Use the correct quantity and unit
Electric charge is measured in coulombs, written C. A positive or negative sign indicates the charge balance; it does not change the unit. The charge +4 μC means +4 × 10⁻⁶ C, while −2 nC means −2 × 10⁻⁹ C. A magnitude comparison uses absolute values separately from the signs.
Milli means 10⁻³, micro means 10⁻⁶ and nano means 10⁻⁹. Write the multiplier before calculating: 6 mC = 6 × 10⁻³ C = 0.006 C. Upper-case C is the coulomb symbol; lower-case c is not a substitute. These conversions support the unit statement without introducing an unprovided current rate or potential difference.
Compare and add with consistent units
Before adding charges, convert them to the same unit and retain both signs. +3 μC and −1 μC sum to +2 μC. Equal and opposite charges can give zero combined charge without making either individual object uncharged. State the system whose total is being calculated.
If an elementary charge magnitude is supplied, a particle imbalance can be converted by multiplying its signed count by that value. For a supplied e = 1.6 × 10⁻¹⁹ C, a deficit of five electrons means +5e = +8.0 × 10⁻¹⁹ C. The unit objective does not require memorising this supplied value, and a macroscopic charge does not imply the tiny illustrative particle counts used in earlier diagrams.
Pause and explain
Which conversion correctly preserves −3 μC?
Worked example
Convert +4 μC and −2 nC to coulombs.
Show the worked solution
- Micro contributes a multiplier of 10⁻⁶.
- Nano contributes a multiplier of 10⁻⁹.
- Keep the signs attached to both converted values.
Answer +4 × 10⁻⁶ C; −2 × 10⁻⁹ C.
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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.
Convert 6 mC and −8 μC to coulombs.
Give me a hint
Attach the supplied sign to the appropriate milli or micro multiplier.
Compare my reasoning
- Milli is 10⁻³: 6 mC = 0.006 C.
- Micro is 10⁻⁶: −8 μC = −0.000008 C.
- Both values retain coulomb units.
0.006 C; −8 × 10⁻⁶ C.
Look for these in your work
- Use the correct powers of ten.
- Keep the negative sign.
Find the combined charge of +3 μC and −1000 nC in coulombs.
Give me a hint
Convert the nano value to the same scale before adding.
Compare my reasoning
- +3 μC = +3 × 10⁻⁶ C.
- −1000 nC = −1 × 10⁻⁶ C.
- Their sum is +2 × 10⁻⁶ C.
+2 × 10⁻⁶ C.
Look for these in your work
- Convert both values.
- Add signed quantities.
An initially neutral model loses five electrons. Use supplied e = 1.6 × 10⁻¹⁹ C to find its charge, then assess a claim that it gained negative charge.
Give me a hint
Electron loss gives a positive imbalance of five elementary units.
Compare my reasoning
- Removing five electrons produces +5e.
- 5 × 1.6 × 10⁻¹⁹ = 8.0 × 10⁻¹⁹ C.
- The negative-gain claim has the wrong sign.
+8.0 × 10⁻¹⁹ C; electron loss gives positive charge.
Look for these in your work
- Use the supplied elementary magnitude.
- Determine the sign independently of the multiplication.
Common mistakes
- Charge uses coulombs, not volts.
- Keep the sign when converting.
- Use one consistent unit before adding.
What can you explain now?
What is the SI unit of electric charge?
Compare with the explanation
Coulomb (C).
The sign of a charge and its magnitude both retain the same coulomb unit.
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Make it stick
State coulombs, convert signed prefixes and check sums without confusing charge with voltage or current.
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.