The big idea: Put a charge in an electric field and it feels 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.
[Diagram: phys-field-lines] - Available in full study mode
[Diagram: phys-free-body] - Available in full study mode
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.
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)
[Diagram: phys-formula-triangle] - Available in full study mode
- 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.
Worked example — acceleration of an electron
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.
Solution
- First the force — start with the given formula:
- Put in the numbers (q = 1.6 × 10⁻¹⁹, E = 2.0 × 10⁴):
- Now the acceleration — use the given Newton's second law:
- Put in the numbers (F = 3.2 × 10⁻¹⁵, m = 9.1 × 10⁻³¹):
- Work it out — keep the unit:
Final answer
a = 3.5 × 10¹⁵ m s⁻² — huge, because the electron's mass is tiny.
Know your predicted grade
Take timed mock exams and get detailed feedback on every answer. See exactly where you're losing marks.
How this is tested: Charged particles between plates are usually an extended Paper 2 calculation in the Fields theme.
- Paper 2: state the direction of the acceleration, then show that the acceleration is of a stated huge order (often 10¹⁴ m s⁻²) using F = qE and a = F ÷ m. - Paper 2: determine how far a particle fired across the gap travels along the plates before it hits one — a projectile problem (constant velocity one way, constant acceleration the other).
Classic trap: treating 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.
IB-style question — (a) the acceleration
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. Find the magnitude of its acceleration.
Solution
- First the force — start with the given formula:
- Put in the numbers (q = 1.6 × 10⁻¹⁹, E = 5.0 × 10³):
- Now the given Newton's second law:
- Put in the numbers (F = 8.0 × 10⁻¹⁶, m = 9.1 × 10⁻³¹):
- Work it out — keep the unit:
Final answer
a = 8.8 × 10¹⁴ m s⁻² — of order 10¹⁴ m s⁻², as expected for a light electron.
IB-style question — (b) the sideways deflection
The same electron spends 2.0 × 10⁻⁹ s crossing the plates. Using the acceleration from part (a), find how far it is deflected sideways (across the field) in that time. It enters moving parallel to the plates, so its sideways speed starts at zero.
Solution
- Across the field the motion is accelerated from rest, so use the suvat equation:
- It starts with no sideways speed (u = 0):
- Put in the numbers (a = 8.8 × 10¹⁴, t = 2.0 × 10⁻⁹):
- Work it out — keep the unit:
Final answer
sideways deflection ≈ 1.8 × 10⁻³ m (about 1.8 mm). Use s = ½at², not s = vt — the sideways motion is accelerated.