A2 · Paper 4 practice · Electric fields

Charge, force, and potential.

A full set of ten original structured questions in the style of Paper 4, covering the whole of Electric fields: field lines, uniform fields, Coulomb’s law, the field of a point charge, and electric potential. Several questions chain one lesson into the next, taking the force between charges through to the field and the potential, all within the topic. Each is linked to its lessons; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards. Take 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻², e = 1.60 × 10⁻¹⁹ C and m₃ = 9.11 × 10⁻³¹ kg where needed.

Keep these straight

Fields, forces, potential.

Uniform field: E = V / d; the force on a charge is F = qE, and a charge crossing the field follows a parabola.
Point charges: Coulomb's law F = Q₁Q₂ / (4πε₀r²) and field E = Q / (4πε₀r²) are both inverse-square.
Potential: V = Q / (4πε₀r) falls off as 1/r; the work to move charge q is W = qΔV, and E = −dV/dr.

Paper 4

[13 marks]

Uniform fields →

Two horizontal parallel plates are 2.0 cm apart with a potential difference of 600 V between them. An electron enters midway between the plates, moving parallel to them at 6.0 × 10⁷ m s⁻¹. The plates are 6.0 cm long. Fig. 1.1 shows the arrangement.

Fig. 1.1

(a) Calculate the electric field strength between the plates. [2]

(b) Calculate the magnitude of the electric force on the electron. [2]

(c) State the direction of this force relative to the field, and explain why. [1]

(d) Explain why the path of the electron between the plates is parabolic. [2]

(e) Calculate the time the electron spends between the plates. [2]

(f) Calculate the vertical deflection of the electron as it leaves the plates. [3]

(g) State the effect on the deflection of doubling the plate voltage. [1]

(a)E = V / d = 600 / 0.020 ✓ = 3.0 × 10⁴ V m⁻¹ ✓
(b)F = qE = 1.60 × 10⁻¹⁹ × 3.0 × 10⁴ ✓ = 4.8 × 10⁻¹⁵ N ✓
(c)Opposite to the field (toward the positive plate), because the electron carries a negative charge ✓
(d)The horizontal velocity is constant while the uniform field gives a constant vertical acceleration ✓; constant velocity in one direction and constant acceleration at right angles to it produce a parabola, just like a projectile ✓
(e)t = L / v = 0.060 / (6.0 × 10⁷) ✓ = 1.0 × 10⁻⁹ s ✓
(f)a = F / m = 4.8 × 10⁻¹⁵ / 9.11 × 10⁻³¹ = 5.3 × 10¹⁵ m s⁻² ✓; y = ½at² = 0.5 × 5.27 × 10¹⁵ × (1.0 × 10⁻⁹)² ✓ = 2.6 × 10⁻³ m ✓
(g)The deflection doubles, since y ∝ a ∝ E ∝ V ✓

Paper 4

[13 marks]

Coulomb’s law →

Two small charged spheres carry charges Q₁ = +2.0 × 10⁻⁸ C and Q₂ = −3.0 × 10⁻⁸ C. Their centres are 4.0 cm apart, as shown in Fig. 2.1.

Fig. 2.1

(a) State Coulomb's law. [2]

(b) Calculate the magnitude of the force between the spheres and state whether it is attractive or repulsive. [3]

(c) State how the force would change if the separation were halved. [2]

(d) Calculate the electric field strength due to Q₁ alone at the position of Q₂. [3]

(e) Hence verify, using F = Q₂E, the force found in (b). [2]

(f) State why a uniformly charged sphere can be treated as a point charge at its centre. [1]

(a)The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation ✓, directed along the line joining them ✓
(b)F = Q₁Q₂ / (4πε₀r²) = 8.99 × 10⁹ × (2.0 × 10⁻⁸)(3.0 × 10⁻⁸) / (0.040)² ✓✓ = 3.4 × 10⁻³ N, attractive (opposite signs) ✓
(c)It would become four times as large ✓, since F ∝ 1/r² ✓
(d)E = Q₁ / (4πε₀r²) = 8.99 × 10⁹ × 2.0 × 10⁻⁸ / (0.040)² ✓✓ = 1.1 × 10⁵ V m⁻¹ ✓
(e)F = Q₂E = 3.0 × 10⁻⁸ × 1.12 × 10⁵ ✓ = 3.4 × 10⁻³ N, the same as (b) ✓
(f)Outside the sphere its field is identical to that of a point charge of the same value placed at its centre ✓

Paper 4

[12 marks]

Electric potential →

An isolated point charge Q = +6.0 × 10⁻⁹ C sits in a vacuum. Fig. 3.1 shows how the electric potential V varies with distance r from the charge.

Fig. 3.1

(a) Define the electric potential at a point. [2]

(b) Calculate the potential at a distance of 3.0 cm from Q. [2]

(c) State the potential at 6.0 cm from Q, explaining your reasoning. [1]

(d) A charge q = +2.0 × 10⁻⁹ C is moved from 6.0 cm to 3.0 cm from Q. Calculate the work done on it. [3]

(e) State the relationship between the electric field strength and the potential. [2]

(f) State how the electric potential energy of two like point charges varies with their separation. [2]

(a)The work done per unit positive charge ✓ in bringing it from infinity to that point ✓
(b)V = Q / (4πε₀r) = 8.99 × 10⁹ × 6.0 × 10⁻⁹ / 0.030 ✓ = 1.8 × 10³ V ✓
(c)9.0 × 10² V: doubling r halves V, because V ∝ 1/r ✓
(d)W = qΔV = q(V₃ − V₆) = 2.0 × 10⁻⁹ × (1798 − 899) ✓✓ = 1.8 × 10⁻⁶ J ✓
(e)The field is the negative gradient of the potential ✓: E = −dV/dr ✓
(f)The potential energy is positive and proportional to 1/r ✓, falling toward zero as the separation increases ✓

Paper 4

[10 marks]

Field lines →

This question is about electric field strength and field lines.

(a) Define electric field strength. [2]

(b) State the equation relating force, charge and field strength. [1]

(c) A charge of +2.0 μC experiences a force of 0.30 N. Calculate the field strength. [2]

(d) State the unit of electric field strength in two equivalent forms. [1]

(e) Describe the field lines of (i) a uniform field and (ii) the radial field of a positive point charge. [2]

(f) State what the direction of a field line shows and what the spacing of the lines shows. [2]

(a)The force per unit positive charge at a point ✓✓
(b)F = qE ✓
(c)E = F/q = 0.30 / 2.0 × 10⁻⁶ ✓ = 1.5 × 10⁵ N C⁻¹ ✓
(d)N C⁻¹ or V m⁻¹ ✓
(e)Uniform: parallel, equally spaced lines ✓; radial: straight lines directed outward from the charge ✓
(f)The direction shows the force on a positive charge ✓; the spacing shows the field strength (closer lines mean a stronger field) ✓

Paper 4

[11 marks]

Uniform fields →

Two parallel plates 4.0 cm apart have a potential difference of 2000 V across them.

(a) Calculate the electric field strength between the plates. [2]

(b) State the direction and nature of the field between the plates. [2]

(c) Calculate the force on an electron between the plates. [2]

(d) Calculate the acceleration of the electron. [2]

(e) Calculate the work done on the electron as it moves from one plate to the other. [2]

(f) State what happens to this energy. [1]

(a)E = V/d = 2000 / 0.040 ✓ = 5.0 × 10⁴ V m⁻¹ ✓
(b)Uniform ✓, directed from the positive plate to the negative plate ✓
(c)F = qE = 1.60 × 10⁻¹⁹ × 5.0 × 10⁴ ✓ = 8.0 × 10⁻¹⁵ N ✓
(d)a = F/m = 8.0 × 10⁻¹⁵ / 9.11 × 10⁻³¹ ✓ = 8.8 × 10¹⁵ m s⁻² ✓
(e)W = qV = 1.60 × 10⁻¹⁹ × 2000 ✓ = 3.2 × 10⁻¹⁶ J ✓
(f)It becomes kinetic energy of the electron ✓

Paper 4

[11 marks]

Uniform fields →

An electron enters the uniform field between two horizontal plates moving horizontally at 3.0 × 10⁷ m s⁻¹, as shown in Fig. 6.1. The field region is 5.0 cm long and the field strength is 2.0 × 10⁴ V m⁻¹.

Fig. 6.1

(a) State the shape of the path in the field and explain why. [2]

(b) Calculate the vertical force and the vertical acceleration of the electron. [3]

(c) Calculate the time the electron spends in the field. [2]

(d) Calculate the vertical deflection as it leaves the field. [3]

(e) State the direction of the deflection. [1]

(a)Parabolic ✓: the horizontal velocity is constant while a constant vertical force gives constant vertical acceleration (like a projectile) ✓
(b)F = qE = 1.60 × 10⁻¹⁹ × 2.0 × 10⁴ = 3.2 × 10⁻¹⁵ N ✓; a = F/m = 3.2 × 10⁻¹⁵ / 9.11 × 10⁻³¹ ✓ = 3.5 × 10¹⁵ m s⁻² ✓
(c)t = L/v = 0.050 / 3.0 × 10⁷ ✓ = 1.7 × 10⁻⁹ s ✓
(d)y = ½at² = ½ × 3.51 × 10¹⁵ × (1.67 × 10⁻⁹)² ✓✓ = 4.9 × 10⁻³ m (about 4.9 mm) ✓
(e)Toward the positive plate ✓

Paper 4

[10 marks]

Coulomb’s law →

Two small charges, +3.0 nC and +5.0 nC, are placed 2.0 cm apart in a vacuum.

(a) State Coulomb’s law. [2]

(b) Calculate the force between the charges. [3]

(c) State whether the force is attractive or repulsive. [1]

(d) State the effect on the force of doubling the separation. [2]

(e) Calculate the separation at which the force would be 1.0 × 10⁻⁴ N. [2]

(a)The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation ✓✓
(b)F = Q₁Q₂/(4πε₀r²) = 8.99 × 10⁹ × (3.0 × 10⁻⁹)(5.0 × 10⁻⁹)/(0.020)² ✓✓ = 3.4 × 10⁻⁴ N ✓
(c)Repulsive (the charges have the same sign) ✓
(d)The force falls to one quarter (inverse-square law) ✓✓
(e)F ∝ 1/r², so r = 0.020 × √(3.37/1.0) ✓ = 3.7 × 10⁻² m ✓

Paper 4

[11 marks]

Coulomb’s law →Field of a point charge →

A point charge Q = +4.0 nC is fixed in a vacuum. This question links Coulomb’s law with the field of a point charge.

(a) A second charge q = +2.0 nC is placed 3.0 cm from Q. Calculate the force on it. [3]

(b) Write the expression for the electric field strength due to a point charge. [1]

(c) Calculate the field strength at 3.0 cm from Q. [2]

(d) Using your answer to (c), show that the force in (a) equals qE. [2]

(e) State and explain how the field strength changes at a distance of 6.0 cm. [2]

(f) State the direction of the field at the point. [1]

(a)F = 8.99 × 10⁹ × (4.0 × 10⁻⁹)(2.0 × 10⁻⁹)/(0.030)² ✓✓ = 8.0 × 10⁻⁵ N ✓
(b)E = Q/(4πε₀r²) ✓
(c)E = 8.99 × 10⁹ × 4.0 × 10⁻⁹/(0.030)² ✓ = 4.0 × 10⁴ V m⁻¹ ✓
(d)qE = 2.0 × 10⁻⁹ × 4.0 × 10⁴ = 8.0 × 10⁻⁵ N ✓, which matches the force in (a) ✓
(e)It falls to one quarter, 1.0 × 10⁴ V m⁻¹ ✓, because E ∝ 1/r² ✓
(f)Radially outward from Q (a positive charge) ✓

Paper 4

[11 marks]

Field of a point charge →Electric potential →

A point charge Q = +6.0 nC is in a vacuum. This question links the field of a point charge with electric potential.

(a) State what is meant by the electric potential at a point. [2]

(b) Calculate the potential at 5.0 cm from Q. [2]

(c) Calculate the field strength at 5.0 cm from Q. [2]

(d) State the relation between E and V for a radial field, and verify it at 5.0 cm. [2]

(e) Calculate the work done in moving a +2.0 nC charge from 5.0 cm to 10.0 cm from Q. [3]

(a)The work done per unit positive charge in bringing it from infinity to that point ✓✓
(b)V = Q/(4πε₀r) = 8.99 × 10⁹ × 6.0 × 10⁻⁹/0.050 ✓ = 1.1 × 10³ V ✓
(c)E = Q/(4πε₀r²) = 8.99 × 10⁹ × 6.0 × 10⁻⁹/(0.050)² ✓ = 2.2 × 10⁴ V m⁻¹ ✓
(d)E = −dV/dr; in magnitude E = V/r for a point charge: 2.16 × 10⁴ × 0.050 = 1.1 × 10³ V = V ✓✓
(e)V at 10 cm = 539 V; ΔV = 539 − 1079 = −540 V ✓; W = qΔV = 2.0 × 10⁻⁹ × (−540) ✓ = −1.1 × 10⁻⁶ J (the field does the work as the like charge moves away) ✓

Paper 4

[12 marks]

Field lines →Coulomb’s law →Field of a point charge →Electric potential →

Two point charges, A = +5.0 nC and B = −5.0 nC, are fixed 8.0 cm apart in a vacuum. P is the midpoint, as shown in Fig. 10.1. This question uses field strength, Coulomb’s law, the field of a point charge, and potential.

Fig. 10.1

(a) Define electric field strength. [1]

(b) Calculate the force between A and B. [2]

(c) Calculate the field strength at P due to each charge, and the resultant field at P. [4]

(d) Calculate the electric potential at P. [3]

(e) State what your answer to (d) implies about the work done in bringing a charge to P from infinity. [2]

(a)The force per unit positive charge at a point ✓
(b)F = 8.99 × 10⁹ × (5.0 × 10⁻⁹)²/(0.080)² ✓ = 3.5 × 10⁻⁵ N (attractive) ✓
(c)Each at r = 0.040 m: E = 8.99 × 10⁹ × 5.0 × 10⁻⁹/(0.040)² = 2.8 × 10⁴ V m⁻¹ ✓✓; at P both fields point from A toward B, so they add: resultant = 5.6 × 10⁴ V m⁻¹ ✓✓
(d)V = ΣQ/(4πε₀r) = 8.99 × 10⁹ × (5.0 × 10⁻⁹/0.040 − 5.0 × 10⁻⁹/0.040) ✓✓ = 0 V ✓
(e)No net work is needed, since the potential at P is zero ✓✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.