The big idea: An inkjet printer steers tiny charged ink drops between two charged plates — and every drop feels the same steady push, wherever it sits in the gap.
A uniform electric field
- That even push is a uniform electric field: put a voltage across two parallel plates and you make one.
- Uniform means the same strength everywhere.
- The field lines are evenly spaced and parallel, running straight from the + plate to the − plate.
A uniform field between parallel plates: the lines are evenly spaced and parallel, pointing from the + plate to the − plate. Evenly spaced means the field is the same strength everywhere between the plates.
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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.
Drawing the field, and a drop held by it
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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)
V = E d. Cover the one you want: E and d side by side → multiply (V = E d); V above d → divide (E = V ÷ d); V above E → divide (d = V ÷ E).
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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)
Millikan's oil drops: charge comes in lumps: Adjust the voltage until a charged oil drop hangs still: then qE = mg, so q = mg/E = mgd/V.
Millikan measured many drops. Every charge was a whole-number multiple of e = 1.6 × 10⁻¹⁹ C: charge is quantised.
Two parallel plates are 0.020 m apart with a potential difference of 600 V between them. Find the electric field strength between them.
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How this is tested — parallel plates are usually a two-part Paper 2 calculation in the Fields theme:
Paper 2 — the field
- Calculate the field E = V/d from the plate voltage and separation, and draw the evenly-spaced field lines.
Paper 2 — the energy
- Find the energy gained by a charge crossing the plates with W = qV, then state it in electronvolts (eV).
The 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)
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. (a) Find the uniform electric field strength between the plates. (b) The alpha particle crosses the full 250 V; find its kinetic energy at the negative plate, then state it in electronvolts. (1 eV = 1.6 × 10⁻¹⁹ J.)
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