Edexcel IAL Physics revision · A2 — Fields
Electric fields
Gravitational fields with the masses swapped for charges — same inverse-square shape, same potential logic, two new twists: charge comes in two signs, and uniform fields between plates behave completely differently from radial ones.
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What the syllabus demands
- —Use Coulomb's law: F = Qq ÷ 4πε₀r²
- —Define electric field strength E = F/Q; use E = Q ÷ 4πε₀r² for point charges
- —Use E = V/d for uniform fields between parallel plates
- —Define electric potential; use V = Q ÷ 4πε₀r
- —Analyse the motion of charged particles in uniform electric fields
Definitions that earn marks
Clear definitions to practise — check your course mark scheme
- Electric field strength
- The force per unit positive charge at a point: E = F ÷ Q. Units: N/C or V/m.
- Electric potential
- The work done per unit positive charge in bringing a small test charge from infinity to the point: V = Q ÷ 4πε₀r.
- Coulomb's law
- The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation.
The equations
More equations to practise: the Edexcel IAL formula sheet.
Where the marks die
Common mistakes to check
- 01
Using E = V/d for a point charge, or the inverse-square formula between plates. Radial fields: inverse-square. Parallel plates: uniform, E = V/d. Identify the geometry before choosing the equation.
- 02
Sign carelessness with potential: unlike gravitational potential, electric potential can be positive (near positive charge) or negative (near negative charge). Keep the sign of Q in the formula.
- 03
Field strength falls as 1/r², but potential falls as 1/r. Mixing the powers is the classic error on sketch-graph questions.
- 04
A charged particle between plates follows a parabola — treat it exactly like projectile motion: constant velocity along the plates, constant acceleration (F = EQ, a = EQ/m) across them.
One worked example, done properly
Question
An electron enters the field between parallel plates 2.0 cm apart with a p.d. of 400 V. Find the force on the electron. (e = 1.6 × 10⁻¹⁹ C)
Method
- 1.E = V/d = 400 ÷ 0.020 = 2.0 × 10⁴ V/m.
- 2.F = EQ = 2.0 × 10⁴ × 1.6 × 10⁻¹⁹.
F = 3.2 × 10⁻¹⁵ N, towards the positive plate
Test yourself
10 questions · instant marking
Question 1 of 10
The electric field strength between parallel plates 5.0 cm apart with a p.d. of 200 V is:
Question 2 of 10
The field between parallel plates (away from the edges) is:
Question 3 of 10
Electric field strength of a point charge falls with distance as:
Question 4 of 10
Doubling the distance from a point charge changes the force on a test charge by a factor of:
Question 5 of 10
A charged particle moving parallel to the plates in a uniform field follows a path that is:
Question 6 of 10
Electric potential near an isolated negative charge is:
Question 7 of 10
The force on an electron (q = 1.6 × 10⁻¹⁹ C) in a field of 5.0 × 10⁴ V/m is:
Question 8 of 10
Unlike gravitational fields, electric fields can:
Question 9 of 10
The work done moving a charge q through a potential difference V is:
Question 10 of 10
Where the equipotential lines around a charge are closest together, the field is: