The big idea: Rub a balloon on your hair and it tugs tiny scraps of paper toward it — those charged scraps sit in the balloon's electric field and feel a force.
The force is F = qE — the charge times the field strength. In a uniform field (like the one between two parallel plates) that force is the same everywhere.
A steady force means a steady acceleration — the charge speeds up in a straight line, or curves if it was already moving.
The uniform field between two charged plates: evenly-spaced parallel lines from the + plate to the − plate. A charge anywhere in this gap feels the same force F = qE.
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A positive charge in a uniform field feels a single steady force F = qE, in the direction of the field (here, downward toward the − plate). Same size everywhere between the plates.
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Spot it: One charge, one field, one force: F = qE, pointing along the field (for a positive charge) or against it (for a negative charge like an electron).
The force is constant, so the acceleration is constant.
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Finding how a charge moves is a two-step chain. First the field gives the force; then Newton's second law turns that force into an acceleration:
- electric force on the charge (N)
- the charge in the field (C, coulombs)
- electric field strength (N C⁻¹, or V m⁻¹)
F = q E. Cover the one you want: q and E side by side → multiply (F = q E); F above q → divide (E = F ÷ q); F above E → divide (q = F ÷ E).
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- net (electric) force on the charge (N)
- mass of the charged particle (kg)
- acceleration of the particle (m s⁻²)
Fired across the field? It's a projectile: If the charge is fired across the gap (sideways to the field), it follows a parabola — the same maths as a thrown ball.
Along the plates: no force, so constant velocity.
Across the plates: constant force, so constant acceleration a = qE ÷ m. The two motions share the same time.
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A uniform field of 2.0 × 10⁴ N C⁻¹ acts on an electron (charge 1.6 × 10⁻¹⁹ C, mass 9.1 × 10⁻³¹ kg). Find the electron's acceleration.
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How this is tested — charged particles between plates are usually an extended Paper 2 calculation in the Fields theme:
Paper 2 — the acceleration
- State the direction of the acceleration (with the field for +, opposite it for −).
- Show that a = qE ÷ m is of a stated huge order — often 10¹⁴ m s⁻².
Paper 2 — the deflection
- Determine how far a particle fired across the gap travels before it hits a plate.
- A projectile: constant velocity along the plates, s = ½at² across them.
The classic trap: Don't treat the sideways deflection as constant-velocity. Across the field the motion is accelerated, so use the suvat equation s = ½at², not s = vt.
The projectile recipe: Along the plates (no force): distance = horizontal speed × time → use this to find the time in the gap.
Across the plates (force qE): the sideways shift is s = ½at², with a = qE ÷ m and the same time. That sideways shift tells you whether — and where — it hits a plate.
An electron (charge 1.6 × 10⁻¹⁹ C, mass 9.1 × 10⁻³¹ kg) enters the uniform field between two parallel plates, where the field strength is 5.0 × 10³ N C⁻¹. (a) Find the magnitude of its acceleration. (b) The electron spends 2.0 × 10⁻⁹ s crossing the plates and enters moving parallel to them, so its sideways speed starts at zero. Find how far it is deflected sideways in that time.
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