A Level Physics 18.5 Electric Potential MCQs – 30 Practice Questions with Answers (Cambridge 9702)

These 30 original practice questions cover Section 18.5, Electric Potential, of the Cambridge International AS & A Level Physics (9702) syllabus, an A Level (A2) topic. They develop the definition of electric potential, potential energy and work calculations, and equipotential surfaces with the potential gradient, using the formula recap below.

Formula recap (full list: A Level Physics Formula Sheet)

  • Potential at a distance r from a point charge Q: V = Q / (4πε0r), where 1 / (4πε0) = 9.0 x 109 m/F
  • Potential energy of a charge q at a point of potential V: Ep = qV
  • Work done moving a charge q through a potential change ΔV: W = qΔV
  • Field strength from the potential gradient: E = −dV/dr (in a uniform field, E = ΔV / Δr = V/d)
  • Electron charge e = 1.6 x 10−19 C; 1 eV = 1.6 x 10−19 J

Electric Potential

Q1. The electric potential at a point in an electric field is defined as the work done per unit

  • (A) mass in bringing a mass from infinity to the point
  • (B) positive charge in bringing a small positive charge from infinity to the point
  • (C) current in bringing a charge from infinity to the point
  • (D) force in bringing a positive charge from infinity to the point
Show Answer

Answer: B – Electric potential is the work done per unit positive charge in bringing a (small test) positive charge from infinity to the point, against the field. It is a property of the point in the field, not of the charge placed there.

Q2. The unit of electric potential, the volt, is equivalent to

  • (A) one newton per coulomb
  • (B) one joule per metre
  • (C) one joule per coulomb
  • (D) one coulomb per joule
Show Answer

Answer: C – Potential is energy (work done) divided by charge, so 1 V = 1 J/C. (One newton per coulomb is the unit of electric field strength.)

Q3. Which statement about electric potential is correct?

  • (A) It is a scalar quantity which may be positive, negative or zero.
  • (B) It is a vector quantity directed away from a positive charge.
  • (C) It is always positive, because work is always done on the charge.
  • (D) It depends on the charge that is brought to the point.
Show Answer

Answer: A – Potential is a scalar: it has magnitude and sign but no direction. Its sign follows the source charge: positive near an isolated positive charge, negative near an isolated negative charge. It does not depend on the test charge placed at the point.

Q4. In calculations of electric potential, the zero of potential is conventionally taken to be at

  • (A) the surface of the Earth
  • (B) the negative terminal of the supply
  • (C) the centre of the source charge
  • (D) infinity
Show Answer

Answer: D – The potential at infinity is defined as zero, which is why V = Q / (4πε0r) is positive near a positive charge and negative near a negative charge.

Q5. A point charge of +2.0 μC is in a vacuum. The electric potential at a point 0.30 m from the charge is

  • (A) 6.0 x 103 V
  • (B) 6.0 x 104 V
  • (C) 2.0 x 105 V
  • (D) 1.8 x 105 V
Show Answer

Answer: B – V = Q / (4πε0r) = (9.0 x 109 x 2.0 x 10−6) / 0.30 = 1.8 x 104 / 0.30 = 6.0 x 104 V.

Q6. The electric potential at a distance of 0.20 m from an isolated point charge of −4.0 μC is

  • (A) +1.8 x 105 V
  • (B) −1.8 x 104 V
  • (C) −1.8 x 105 V
  • (D) −3.6 x 104 V
Show Answer

Answer: C – V = (9.0 x 109 x (−4.0 x 10−6)) / 0.20 = −3.6 x 104 / 0.20 = −1.8 x 105 V. Work must be done to bring a positive charge in from infinity against the attraction, so the potential near a negative charge is negative.

Q7. The potential at a point due to a point charge of +1.0 μC is +900 V. The distance of the point from the charge is

  • (A) 10 m
  • (B) 1.0 m
  • (C) 0.10 m
  • (D) 100 m
Show Answer

Answer: A – Rearranging V = Q / (4πε0r): r = (9.0 x 109 x 1.0 x 10−6) / 900 = 9.0 x 103 / 900 = 10 m.

Q8. A point P is 0.10 m from a charge of +3.0 μC and 0.30 m from a charge of −1.0 μC. The resultant potential at P is

  • (A) 3.0 x 105 V
  • (B) 2.7 x 105 V
  • (C) 2.4 x 104 V
  • (D) 2.4 x 105 V
Show Answer

Answer: D – Potential is a scalar, so the contributions are simply added with their signs. From the +3.0 μC charge: V = (9.0 x 109 x 3.0 x 10−6) / 0.10 = +2.7 x 105 V. From the −1.0 μC charge: V = (9.0 x 109 x (−1.0 x 10−6)) / 0.30 = −3.0 x 104 V. Total = 2.7 x 105 − 0.3 x 105 = 2.4 x 105 V.

Q9. The distance from a point charge is doubled. The electric potential at the new point, compared with the original point, is

  • (A) one quarter as large
  • (B) one half as large
  • (C) unchanged
  • (D) twice as large
Show Answer

Answer: B – V is proportional to 1/r (not 1/r2), so doubling r halves V. The 1/r2 dependence belongs to the field strength E of a point charge.

Q10. The electric potential at a distance of 1.0 x 10−10 m from a proton (charge +1.6 x 10−19 C) is approximately

  • (A) 1.4 V
  • (B) 144 V
  • (C) 14 V
  • (D) 0.14 V
Show Answer

Answer: C – V = (9.0 x 109 x 1.6 x 10−19) / (1.0 x 10−10) = 1.44 x 10−9 / 1.0 x 10−10 = 14.4 V, approximately 14 V.

Potential Energy & Work

Q11. The electric potential energy of a charge q placed at a point where the potential is V is given by

  • (A) q / V
  • (B) V / q
  • (C) qV
  • (D) qV2
Show Answer

Answer: C – Ep = qV. Because q may be negative, the potential energy of a negative charge at a point of positive potential is negative.

Q12. The work done by an external agent in moving a charge of +2.0 μC from a point at potential +100 V to a point at potential +250 V is

  • (A) 3.0 x 10−4 J
  • (B) 5.0 x 10−4 J
  • (C) 2.0 x 10−4 J
  • (D) 7.0 x 10−4 J
Show Answer

Answer: A – W = qΔV = 2.0 x 10−6 x (250 − 100) = 2.0 x 10−6 x 150 = 3.0 x 10−4 J. Only the potential difference matters, not the starting potential alone.

Q13. A charge is moved between two points in an electric field along two different routes. The work done is the same for both routes because

  • (A) the field is uniform along both routes
  • (B) the work done depends only on the charge and the potential difference between the endpoints
  • (C) no force acts on the charge during the journey
  • (D) the charge moves at constant speed on both routes
Show Answer

Answer: B – The electric field is conservative: W = q(V2 − V1) depends only on the potentials at the start and finish, so it is independent of the path taken.

Q14. A small positive charge is released from rest in an electric field. It moves

  • (A) towards lower potential, gaining potential energy
  • (B) towards higher potential, losing kinetic energy
  • (C) along an equipotential surface, at constant energy
  • (D) towards lower potential, losing potential energy and gaining kinetic energy
Show Answer

Answer: D – The field does work on a positive charge in the direction of the force, which is from high potential to low potential. Its potential energy qV decreases and the lost potential energy becomes kinetic energy.

Q15. An electron is released from rest in an electric field. It moves towards a point of

  • (A) lower potential, and its potential energy increases
  • (B) higher potential, and its potential energy increases
  • (C) lower potential, and its potential energy decreases
  • (D) higher potential, and its potential energy decreases
Show Answer

Answer: D – The force on the negatively charged electron is opposite to the field direction, so it moves from low potential to high potential. Its potential energy Ep = qV = (−e)V decreases as V increases, and the lost potential energy becomes kinetic energy.

Q16. A proton, initially at rest, is accelerated through a potential difference of 500 V. The kinetic energy it gains is

  • (A) 8.0 x 10−17 J
  • (B) 8.0 x 10−19 J
  • (C) 3.2 x 10−16 J
  • (D) 500 J
Show Answer

Answer: A – Gain in KE = qΔV = 1.6 x 10−19 x 500 = 8.0 x 10−17 J, i.e. 500 eV.

Q17. An electron (mass 9.11 x 10−31 kg) is accelerated from rest through a potential difference of 100 V. Its final speed is approximately

  • (A) 5.9 x 105 m/s
  • (B) 1.9 x 106 m/s
  • (C) 5.9 x 106 m/s
  • (D) 3.5 x 107 m/s
Show Answer

Answer: C – eV = 0.5 m v2, so v = sqrt(2 x 1.6 x 10−19 x 100 / 9.11 x 10−31) = sqrt(3.5 x 1013) = 5.9 x 106 m/s.

Q18. A charge of −2.0 μC is placed at a point where the potential is +400 V. Its electric potential energy is

  • (A) +8.0 x 10−4 J
  • (B) +2.0 x 10−4 J
  • (C) −2.0 x 10−4 J
  • (D) −8.0 x 10−4 J
Show Answer

Answer: D – Ep = qV = (−2.0 x 10−6) x (+400) = −8.0 x 10−4 J. The negative sign shows the charge is bound: work would have to be done to remove it to infinity.

Q19. A proton moves, under the action of the field alone, from a point at potential +800 V to a point at potential +200 V. The gain in its kinetic energy is

  • (A) 1.6 x 10−16 J
  • (B) 9.6 x 10−17 J
  • (C) 6.0 x 10−17 J
  • (D) 1.0 x 10−16 J
Show Answer

Answer: B – Loss of potential energy = q(V1 − V2) = 1.6 x 10−19 x (800 − 200) = 1.6 x 10−19 x 600 = 9.6 x 10−17 J, and this becomes kinetic energy.

Q20. One electronvolt (1 eV) is the energy gained by

  • (A) an electron accelerated through a potential difference of 1 V, equal to 1.6 x 10−19 J
  • (B) a proton accelerated through a potential difference of 1 V, equal to 1 J
  • (C) any charge accelerated through a potential difference of 1 V, equal to 1.6 x 10−19 J
  • (D) an electron accelerated through a potential difference of 1 kV, equal to 1.6 x 10−19 J
Show Answer

Answer: A – 1 eV = e x 1 V = 1.6 x 10−19 C x 1 V = 1.6 x 10−19 J. It applies to any particle carrying a charge of magnitude e, such as a proton, moving through 1 V.


Equipotentials & Potential Gradient

Q21. The equipotential surfaces around an isolated point charge are

  • (A) parallel planes
  • (B) concentric spheres centred on the charge
  • (C) radial lines from the charge
  • (D) concentric cylinders around the charge
Show Answer

Answer: B – V depends only on the distance r from the charge, so all points at the same r have the same potential: the equipotentials are spheres centred on the charge. (Radial lines are the field lines, not the equipotentials.)

Q22. A charge is moved from one point to another along the same equipotential surface. The work done on the charge is

  • (A) equal to q times the potential of the surface
  • (B) equal to the charge times the field strength
  • (C) zero
  • (D) at its maximum, because the path is longest
Show Answer

Answer: C – W = qΔV and ΔV = 0 between points at the same potential, so no work is done. This is why the field (and its force) is always perpendicular to an equipotential surface.

Q23. At any point, an equipotential surface and the electric field line through that point are

  • (A) perpendicular to each other
  • (B) parallel to each other
  • (C) at 45 degrees to each other
  • (D) in the same direction
Show Answer

Answer: A – If the field had a component along the equipotential surface, work would be done moving a charge along it, which contradicts ΔV = 0. Field lines therefore cross equipotentials at right angles.

Q24. Equipotential surfaces are drawn for equal intervals of potential. Where the surfaces are closest together, the electric field is

  • (A) zero
  • (B) weakest
  • (C) uniform
  • (D) strongest
Show Answer

Answer: D – E = −dV/dr: closely spaced equipotentials mean the potential changes rapidly with distance, so the potential gradient, and hence the field strength, is large there.

Q25. In a uniform electric field, such as that between parallel plates, the equipotential surfaces are

  • (A) spheres centred midway between the plates
  • (B) planes parallel to the plates, perpendicular to the field lines
  • (C) planes perpendicular to the plates
  • (D) lines parallel to the plates
Show Answer

Answer: B – In a uniform field the potential changes steadily along the field direction only, so surfaces of constant potential are planes parallel to the plates and at right angles to the (parallel) field lines.

Q26. In a uniform electric field, the potential falls from 500 V to 100 V over a distance of 0.20 m measured along the field. The field strength is

  • (A) 400 V/m
  • (B) 80 V/m
  • (C) 2000 V/m
  • (D) 3000 V/m
Show Answer

Answer: C – E = ΔV / Δr = (500 − 100) / 0.20 = 400 / 0.20 = 2000 V/m (equivalently 2000 N/C).

Q27. The relation E = −dV/dr contains a negative sign because

  • (A) the field points in the direction of decreasing potential
  • (B) potential is always negative
  • (C) field strength is always negative
  • (D) charge is measured from infinity
Show Answer

Answer: A – Potential decreases in the direction of the field (a positive charge loses potential energy as it moves with the field). The negative sign in E = −dV/dr ensures E is positive in the direction of falling V.

Q28. The potential gradient at a point in an electric field may be measured in

  • (A) volts
  • (B) volts per metre
  • (C) joules per coulomb
  • (D) newtons per coulomb per metre
Show Answer

Answer: B – The gradient is ΔV / Δr, potential change per unit distance, so its unit is V/m – the same unit as electric field strength, since E equals the negative of the gradient.

Q29. A key difference between gravitational potential and electric potential is that gravitational potential

  • (A) is a vector, whereas electric potential is a scalar
  • (B) is measured in J/C, whereas electric potential is measured in J/kg
  • (C) increases towards a mass, whereas electric potential decreases towards a positive charge
  • (D) is always negative, whereas electric potential may be positive or negative
Show Answer

Answer: D – Mass only attracts, so gravitational potential (in J/kg) is always negative, reaching zero at infinity. Charges may be positive or negative, so electric potential (in J/C) can have either sign. Both quantities are scalars defined with zero at infinity.

Q30. Two parallel plates, 3.0 cm apart, are connected to a 240 V supply, one plate being at 0 V and the other at +240 V. The potential at a point 1.0 cm from the 0 V plate, measured along the field, is

  • (A) 24 V
  • (B) 120 V
  • (C) 80 V
  • (D) 160 V
Show Answer

Answer: C – The field is uniform, so the potential rises linearly from 0 V to 240 V across 3.0 cm. At 1.0 cm, one third of the way across: V = 240 x (1.0 / 3.0) = 80 V.

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