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.

Verified against AQA 7408 (2026 spec)

Know the equation

SymbolQuantityUnit
Eelectric field strengthN/C
Qcharge producing the fieldC
epsilon0permittivity of free spaceF/m
rdistance from the point chargem
Velectric potentialV = J/C
qcharge moved in the fieldC

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
3 marks

Calculate the electric potential 0.15 m from a point charge of +3.2 nC. Use epsilon0 = 8.85 x 10−12 F/m.

Do the calculation on paper first — then mark it.

Q2
3 marks

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.

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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