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Force per unit positive charge.

Six original Cambridge-style questions on the definition of electric field strength, the force F = qE on positive and negative charges, and what field lines show.

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

Define it with a positive test charge.

01
Calculation
[2 marks]

At a point in air the electric field strength is 2.5 × 10⁴ N C⁻¹, directed vertically downwards. Calculate the electric force on an electron placed at that point, and state its direction.

  • F = qE = 1.60 × 10⁻¹⁹ × 2.5 × 10⁴ = 4.0 × 10⁻¹⁵ N. ✓
  • The force is vertically upwards: field strength is defined by the force on a positive charge, and the electron's charge is negative, so its force is opposite to the field. ✓
02
Analysis
[3 marks]

Two equal positive point charges are fixed a short distance apart. Describe the pattern of electric field lines around them, and state what happens to the field at the point midway between the charges.

  • Field lines leave each charge radially, pointing away from it, because a positive test charge would be repelled by both. ✓
  • Lines from the two charges curve away from each other in the region between them, and far away the pattern looks like the radial field of a single charge of twice the size. ✓
  • At the midpoint the two fields are equal in size and opposite in direction, so the resultant field is zero; no field line passes through this neutral point, and no lines cross anywhere since the field has one direction at each point. ✓
03
Calculation
[3 marks]

A tiny charged droplet of mass 1.47 × 10⁻¹⁴ kg is held stationary in air by a uniform vertical electric field of strength 3.0 × 10⁵ N C⁻¹. Calculate the charge on the droplet and the number of excess electrons it carries. State the direction of the field.

  • For the droplet to be at rest the electric force balances the weight: qE = mg, so q = mg / E = 1.47 × 10⁻¹⁴ × 9.81 / (3.0 × 10⁵). ✓
  • q = 4.8 × 10⁻¹⁹ C, which is 4.81 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3 electron charges, so the droplet carries 3 excess electrons. ✓
  • The electric force on the negative droplet must be upwards, and a negative charge is pushed against the field, so the field is directed vertically downwards. ✓
04
Analysis
[2 marks]

Explain why electric field lines always meet the surface of a charged metal conductor at right angles when the charges on the conductor are at rest.

  • If the field at the surface had a component parallel to the surface, it would exert a force along the surface on the free electrons in the metal, which would move. ✓
  • The charges redistribute until that parallel component is zero, so once they are at rest the only field at the surface is perpendicular to it, and the lines leave or enter at right angles. ✓
05
Calculation
[3 marks]

A proton and an electron are released from rest in the same uniform electric field of strength 1.2 × 10⁴ N C⁻¹. Compare the forces on the two particles, and calculate the initial acceleration of each.

  • Both carry charges of magnitude e, so the forces are equal in size, F = eE = 1.60 × 10⁻¹⁹ × 1.2 × 10⁴ = 1.9 × 10⁻¹⁵ N, but opposite in direction: the proton along the field, the electron against it. ✓
  • Electron: a = F / m = 1.92 × 10⁻¹⁵ / 9.11 × 10⁻³¹ = 2.1 × 10¹⁵ m s⁻². ✓
  • Proton: a = 1.92 × 10⁻¹⁵ / 1.67 × 10⁻²⁷ = 1.1 × 10¹² m s⁻², about 1800 times smaller because the proton is about 1800 times more massive. ✓
06
Calculation
[4 marks]

A small polystyrene ball of mass 0.24 g hangs from an insulating thread in a horizontal uniform electric field of strength 4.5 × 10⁴ N C⁻¹. The ball is charged, and in equilibrium the thread makes an angle of 15° with the vertical.

(a) Write down the two equations obtained by resolving the forces on the ball horizontally and vertically. (b) Calculate the charge on the ball. (c) Calculate the tension in the thread.

  • (a) Horizontally T sin 15° = qE and vertically T cos 15° = mg, where T is the tension. ✓
  • (b) Dividing, tan 15° = qE / mg, so q = mg tan 15° / E = 0.24 × 10⁻³ × 9.81 × 0.268 / (4.5 × 10⁴). ✓
  • q = 1.4 × 10⁻⁸ C (14 nC). ✓
  • (c) T = mg / cos 15° = 0.24 × 10⁻³ × 9.81 / 0.966 = 2.4 × 10⁻³ N, slightly more than the weight. ✓

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.