D.3 · IB Physics
Charge between parallel plates
A charged particle enters the uniform field between two parallel plates, moving parallel to them. The constant force F = qE makes its path a parabola, like a projectile, and a straight line once it leaves the plates (edge effects ignored).
Predict first
Start from the default settings (electron, 100 V, d = 2.0 cm). Keeping V the same, you double the plate separation d to 4.0 cm. What happens to the deflection at the exit?
E = V ÷ d, so doubling d halves E and the acceleration a = qE ÷ m. The time in the field is unchanged, so y = ½at² halves.
Check it in the model: Move the Plate separation, d slider from 2.0 to 4.0 cm and compare the Deflection at exit readout (7.63 mm becomes 3.81 mm).
path of the particle · dashed: previous shot · force on the particle
t = L ÷ v (along the plates)
y = ½at² ; v_y = at
tan θ = v_y ÷ v
ΔE_k = |q|E y
Try this
- Fire an electron at the default settings. Then double the entry speed from 12 to 24 × 10⁶ m s⁻¹ and compare the deflection, the time in the field and the exit angle.
- Press Wider gap: the dashed path is the 2.0 cm shot. Explain the difference using E = V ÷ d, not V.
- Choose the proton at 0.4 × 10⁶ m s⁻¹ and compare its deflection with the electron’s. Use the minimum-speed readout to find how slow the electron can go before it hits a plate.
Common error: using the p.d. V in place of the field strength E in F = qE: the force depends on V ÷ d, so moving the plates apart at the same p.d. reduces the deflection. Also remember that a negative charge accelerates towards the positive plate, against the field direction.