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A2 · Practice questions · Electric potential

Work per unit charge, from infinity.

Six original Cambridge-style questions on the definition of electric potential, V = Q / (4πε₀r), the work W = qΔV, equipotentials and the link E = −dV/dr.

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

Potential is a scalar; the field is its gradient.

01
Analysis
[2 marks]

Define the electric potential at a point, and explain why the potential at every point near an isolated negative charge is negative.

  • The electric potential at a point is the work done per unit positive charge in bringing a small positive test charge from infinity to that point. ✓
  • A positive test charge is attracted towards a negative charge, so the field does work on it as it comes in from infinity and the work done by an external agent is negative; with the potential taken as zero at infinity, the potential is therefore negative everywhere near the charge. ✓
02
Calculation
[3 marks]

A point charge of +5.0 nC is fixed in a vacuum. Calculate the electric potential at a distance of 40 mm from the charge. An electron is moved from this point to a point 80 mm from the charge. Calculate the work that must be done on the electron, and explain its sign.

  • V = Q / (4πε₀r) = 8.99 × 10⁹ × 5.0 × 10⁻⁹ / 0.040 = 1.1 × 10³ V. ✓
  • At 80 mm the potential is half as large, 562 V, so ΔV = 562 − 1124 = −562 V and W = qΔV = (−1.60 × 10⁻¹⁹) × (−562) = 9.0 × 10⁻¹⁷ J. ✓
  • The work done on the electron is positive: the negative electron is attracted to the positive charge, so an external agent must do work to pull it further away, and its electric potential energy rises. ✓
03
Calculation
[3 marks]

Point charges of +6.0 nC and −4.0 nC are fixed 70 mm apart in a vacuum. Point P lies on the line joining them, 30 mm from the +6.0 nC charge and 40 mm from the −4.0 nC charge. Calculate the electric potential at P, and determine the position of the point on the line between the charges at which the potential is zero.

  • Potential is a scalar, so the contributions add with their signs: VP = 8.99 × 10⁹ × (6.0 × 10⁻⁹ / 0.030 − 4.0 × 10⁻⁹ / 0.040) = 8.99 × 10⁹ × 1.0 × 10⁻⁷ = 9.0 × 10² V. ✓
  • At a distance x from the +6.0 nC charge the potential is zero when 6.0 / x = 4.0 / (0.070 − x). ✓
  • 6.0(0.070 − x) = 4.0x gives 0.42 = 10x, so x = 0.042 m: the potential is zero 42 mm from the +6.0 nC charge and 28 mm from the −4.0 nC charge. ✓
04
Analysis
[3 marks]

Explain why no work is done when a charge is moved along an equipotential surface, and hence explain why electric field lines always cross equipotential surfaces at right angles and point from high potential to low potential.

  • The work done moving a charge q is W = qΔV, and along an equipotential ΔV = 0, so no work is done however far the charge moves on the surface. ✓
  • If the field had a component along the surface, the electric force would do work on a charge moving along it, contradicting W = 0; so the field has no component along the surface and must be perpendicular to it. ✓
  • E = −dV/dr: the field equals the negative of the potential gradient, so it points in the direction in which the potential falls most steeply, from high to low potential. ✓
05
Calculation
[3 marks]

An alpha particle of kinetic energy 4.8 MeV is fired directly towards a gold nucleus (proton number 79), which may be assumed to stay at rest. Calculate the distance of closest approach of the alpha particle to the centre of the nucleus.

  • Kinetic energy = 4.8 × 10⁶ × 1.60 × 10⁻¹⁹ = 7.7 × 10⁻¹³ J; at closest approach all of it has become electric potential energy, Qq / (4πε₀r). ✓
  • With Q = 79e and q = 2e: r = 8.99 × 10⁹ × 79 × 1.60 × 10⁻¹⁹ × 2 × 1.60 × 10⁻¹⁹ / 7.68 × 10⁻¹³. ✓
  • r = 4.7 × 10⁻¹⁴ m, a few times the radius of the gold nucleus, so the alpha particle never touches it. ✓
06
Calculation
[4 marks]

A point charge of +2.0 μC is fixed in a vacuum.

(a) Calculate the electric potential at a distance of 0.50 m from the charge. (b) Calculate the electric potential at a distance of 2.0 m from the charge. (c) A charge of +0.50 μC is moved from the point 2.0 m away to the point 0.50 m away. Calculate the work done on it, and state what happens to this energy if the charge is then released.

  • (a) V = Q / (4πε₀r) = 8.99 × 10⁹ × 2.0 × 10⁻⁶ / 0.50 = 3.6 × 10⁴ V. ✓
  • (b) At four times the distance the potential is one quarter: 8.99 × 10⁹ × 2.0 × 10⁻⁶ / 2.0 = 9.0 × 10³ V. ✓
  • (c) W = qΔV = 0.50 × 10⁻⁶ × (3.60 × 10⁴ − 0.90 × 10⁴) = 1.3 × 10⁻² J of work must be done on the charge, since it is pushed against the repulsion. ✓
  • Released, the charge is repelled back outwards and the 1.3 × 10⁻² J of electric potential energy is converted to its kinetic energy by the time it is 2.0 m away. ✓

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.