Calculate the electric potential 0.15 m from a point charge of +3.2 nC. Use epsilon0 = 8.85 x 10−12 F/m.
[3 marks]
Total for this question: 3
All equation deep-dives
Electric field and potential
Fields and capacitors
AQA 7408
In the data booklet
ORIGINAL
E = Q / (4 pi epsilon0 r2) and V = Q / (4 pi epsilon0 r)
Electric potential carries the sign of the source charge; field strength calculations often ask for magnitude only.
Board and spec code confirmed against AQA 7408 · registry checked 2026-07-11How this checking works
Know the equation
| Symbol | Quantity | Unit |
|---|---|---|
| E | electric field strength | N/C |
| Q | charge producing the field | C |
| epsilon0 | permittivity of free space | F/m |
| r | distance from the point charge | m |
| V | electric potential | V = J/C |
| q | charge moved in the field | C |
Rearrangements
- Q = 4 pi epsilon0 E r2
- r = sqrt(Q / (4 pi epsilon0 E))
- Q = 4 pi epsilon0 V r
- d(W) = q d(V)
Apply it — mark your own working
Work each one out on paper first, then reveal the mark scheme and tick the marks you actually earned. That is exactly how you should mark past papers.
Q1
Do the calculation on paper first — then mark it.
Q2
Calculate the magnitude of the electric field strength 0.20 m from a point charge of +6.0 nC. Use epsilon0 = 8.85 x 10−12 F/m.
[3 marks]
Total for this question: 3
Do the calculation on paper first — then mark it.
Where the marks get lost
- Using 1/r for field strength or 1/r2 for potential.
- Leaving nanocoulombs unconverted: 1 nC = 1 x 10−9 C.
- Treating electric potential as always positive; its sign follows the source charge.
Exam tip: Write k = 1/(4 pi epsilon0) if several radial-field calculations are linked, but use the full booklet form in your first line so the method is unambiguous.
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