The big idea: Put a voltage across two parallel metal plates and you make a uniform electric field in the gap between them.
Uniform means it has the same strength everywhere — the field lines are evenly spaced and parallel, running straight from the + plate to the − plate.
Any charge in the gap feels the same steady force, wherever it sits.
[Diagram: phys-field-lines] - Available in full study mode
Spot it: Uniform field = evenly-spaced, parallel field lines.
The lines point from + to − (the direction a positive charge would be pushed). The field is the same strength all the way across the gap.
The strength of the uniform field depends on the voltage across the plates and the gap between them. The bigger the voltage, or the smaller the gap, the stronger the field:
- electric field strength between the plates (V m⁻¹, or N C⁻¹)
- potential difference (voltage) between the plates (V)
- separation (gap) between the plates (m)
[Diagram: phys-formula-triangle] - Available in full study mode
The force on a charge: A charge q placed in the field feels a force given by the field-strength definition (also in the booklet):
F = qE — the bigger the charge or the stronger the field, the bigger the push.
- electric force on the charge (N)
- the charge in the field (C, coulombs)
- electric field strength (N C⁻¹, or V m⁻¹)
Worked example — field between two plates
Two parallel plates are 0.020 m apart with a potential difference of 600 V between them. Find the electric field strength between them.
Solution
- Start with the given formula:
- Put in the numbers (V = 600, d = 0.020):
- Work it out — keep the unit:
Final answer
E = 3.0 × 10⁴ V m⁻¹ — the same value everywhere between the plates.
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How this is tested: Parallel plates are usually a two-part Paper 2 calculation in the Fields theme.
- Paper 2: calculate the field E = V/d from the plate voltage and separation, and draw the evenly-spaced field lines. - Paper 2: find the energy gained by a charge crossing the plates with W = qV, then state it in electronvolts (eV).
Classic trap: forgetting to convert the gap to metres, or mixing up the gap d with the plate length.
Work done moving a charge: W = qV: Moving a charge q through a potential difference V does work on it:
W = qV (energy in joules). For a charge released from rest, this work becomes its kinetic energy.
This one is not in the data booklet — remember it. It is the very meaning of voltage: energy per unit charge.
- work done moving the charge (J, joules) — its energy gain
- the charge moved (C, coulombs)
- potential difference moved through (V, volts)
IB-style question — (a) the field between the plates
An alpha particle of charge +3.2 × 10⁻¹⁹ C is released from the positive plate. The plates are 0.050 m apart with a potential difference of 250 V. Find the uniform electric field strength between the plates.
Solution
- Start with the given formula:
- Put in the numbers (V = 250, d = 0.050):
- Work it out — keep the unit:
Final answer
E = 5.0 × 10³ V m⁻¹ — uniform, so the same everywhere in the gap.
IB-style question — (b) the kinetic energy in eV
The same alpha particle (charge +3.2 × 10⁻¹⁹ C) crosses the full 250 V between the plates. Find its kinetic energy when it reaches the negative plate, then state that energy in electronvolts. (1 eV = 1.6 × 10⁻¹⁹ J.)
Solution
- The work done on the charge becomes its kinetic energy — start with:
- Put in the numbers (q = 3.2 × 10⁻¹⁹, V = 250):
- Work it out:
- Convert to electronvolts (divide by 1.6 × 10⁻¹⁹ J per eV):
Final answer
kinetic energy = 8.0 × 10⁻¹⁷ J = 500 eV. (A charge of 2e crossing 250 V gains 2 × 250 = 500 eV — a quick check.)