A2 · Practice questions · Uniform electric fields

The same field everywhere.

Six original Cambridge-style questions on the uniform field E = V / d between parallel plates, charges accelerated across it, and the parabolic path of a charge fired through it.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Parallel plates: E = V / d, then treat it as a projectile.

01
Calculation
[2 marks]

Two parallel metal plates are 12 mm apart and have a potential difference of 450 V between them. Calculate the electric field strength between the plates and the force on a proton placed between them.

  • E = V / d = 450 / 0.012 = 3.8 × 10⁴ V m⁻¹. ✓
  • F = qE = 1.60 × 10⁻¹⁹ × 3.75 × 10⁴ = 6.0 × 10⁻¹⁵ N, directed from the positive plate towards the negative plate. ✓
02
Analysis
[2 marks]

Explain what is meant by the statement that the electric field between two parallel charged plates is uniform, and describe how this is shown in a diagram of the field lines.

  • The field strength has the same magnitude and the same direction at every point between the plates (away from the edges), so a charge feels the same force wherever it is placed. ✓
  • The field lines are drawn straight, parallel, equally spaced and perpendicular to the plates, running from the positive plate to the negative plate. ✓
03
Calculation
[3 marks]

An electron is released from rest at the negative plate of a pair of parallel plates 2.0 cm apart with a potential difference of 800 V between them. Calculate the kinetic energy and the speed of the electron as it reaches the positive plate, and the time it takes to cross the gap.

  • KE gained = qV = 1.60 × 10⁻¹⁹ × 800 = 1.3 × 10⁻¹⁶ J. ✓
  • ½mv² = 1.28 × 10⁻¹⁶, so v = √(2 × 1.28 × 10⁻¹⁶ / 9.11 × 10⁻³¹) = 1.7 × 10⁷ m s⁻¹. ✓
  • E = 800 / 0.020 = 4.0 × 10⁴ V m⁻¹ and a = eE / m = 7.0 × 10¹⁵ m s⁻²; from d = ½at², t = √(2 × 0.020 / 7.03 × 10¹⁵) = 2.4 × 10⁻⁹ s. ✓
04
Calculation
[3 marks]

An electron travelling at 2.5 × 10⁷ m s⁻¹ enters the uniform field between two horizontal plates, midway between them and parallel to them. The plates are 50 mm long and 20 mm apart, and the potential difference between them is 120 V. Determine whether the electron leaves the field without striking a plate, and calculate the angle its velocity makes with its original direction as it leaves.

  • E = V / d = 120 / 0.020 = 6.0 × 10³ V m⁻¹, so a = eE / m = 1.60 × 10⁻¹⁹ × 6.0 × 10³ / 9.11 × 10⁻³¹ = 1.05 × 10¹⁵ m s⁻² towards the positive plate. ✓
  • Time in the field t = L / v = 0.050 / 2.5 × 10⁷ = 2.0 × 10⁻⁹ s, so the deflection is y = ½at² = 0.5 × 1.05 × 10¹⁵ × (2.0 × 10⁻⁹)² = 2.1 mm, less than the 10 mm to the plate, so the electron leaves the field. ✓
  • vy = at = 1.05 × 10¹⁵ × 2.0 × 10⁻⁹ = 2.1 × 10⁶ m s⁻¹, so tan θ = 2.1 × 10⁶ / 2.5 × 10⁷ and θ = 4.8°. ✓
05
Analysis
[3 marks]

An electron and a proton enter the same uniform electric field at the same point with the same velocity, perpendicular to the field. Compare the forces on them and the paths they follow while in the field.

  • The forces are equal in magnitude, eE, but opposite in direction: the proton is pushed along the field and the electron against it, so they curve towards opposite plates. ✓
  • Both follow parabolic paths, because each keeps a constant velocity along the plates while having a constant acceleration across them. ✓
  • The proton's acceleration eE / m is about 1800 times smaller than the electron's, so in the same time in the field its deflection is about 1800 times smaller and its path is almost straight. ✓
06
Calculation
[4 marks]

In an oil-drop experiment a drop of mass 2.4 × 10⁻¹⁵ kg is held stationary between two horizontal plates 5.0 mm apart when the potential difference between them is 245 V, with the upper plate positive.

(a) Calculate the electric field strength between the plates. (b) Calculate the charge on the drop. (c) State the sign of the charge and the number of electrons it corresponds to.

  • (a) E = V / d = 245 / 5.0 × 10⁻³ = 4.9 × 10⁴ V m⁻¹. ✓
  • (b) The electric force balances the weight, qE = mg, so q = mg / E = 2.4 × 10⁻¹⁵ × 9.81 / 4.9 × 10⁴. ✓
  • q = 4.8 × 10⁻¹⁹ C. ✓
  • (c) The upward force on the drop is towards the positive plate, against the field, so the charge is negative: 4.8 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3 excess electrons. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.