A Level Physics 18.2 Uniform Electric Fields MCQs – 30 Practice Questions with Answers (Cambridge 9702)

Uniform electric fields are at the heart of Cambridge 9702 A Level Section 18.2, covering the field between charged parallel plates, the relation E = V/d, and the motion of charged particles through the field. These 30 original practice questions cover plate fields, electron deflection and applications such as the cathode-ray oscilloscope and Millikan’s oil-drop experiment, each with a fully worked answer.

Formula recap: E = F/Q  |  between parallel plates, E = V/d  |  force on a charge, F = QE  |  acceleration, a = QE/m  |  kinetic energy gained crossing a p.d., ½mv2 = QV  |  Millikan balance, QE = mg. See the full A Level Physics Formula Sheet for every equation used in this topic.

The Field Between Parallel Plates

Q1. Which statement best describes the electric field between two oppositely charged parallel metal plates, away from the edges?

  • (A) The field strength decreases steadily from the positive plate to the negative plate
  • (B) The field is zero at every point between the plates
  • (C) The field is uniform: it has the same magnitude and direction at every point between the plates
  • (D) The field spreads out radially from the centre of the positive plate
Show Answer

Answer: C – Between parallel plates the field lines are parallel and equally spaced, so the field strength E is the same everywhere between the plates (edge effects are ignored). This is what “uniform field” means.

Q2. Two parallel plates are connected to a 200 V supply and are separated by 2.0 cm. What is the electric field strength between the plates?

  • (A) 4.0 V m−1
  • (B) 1.0 × 104 V m−1
  • (C) 1.0 × 102 V m−1
  • (D) 4.0 × 103 V m−1
Show Answer

Answer: B – E = V/d = 200 / 0.020 = 10 000 = 1.0 × 104 V m−1. Remember to convert the separation to metres first.

Q3. The separation of a pair of parallel plates is halved while the potential difference between them is kept the same. What happens to the field strength between the plates?

  • (A) It is halved
  • (B) It is unchanged
  • (C) It is doubled
  • (D) It is quadrupled
Show Answer

Answer: C – E = V/d. With V fixed and d halved, E doubles. Bringing the plates closer together at the same p.d. makes the field stronger.

Q4. A charge of +3.2 × 10−19 C is placed in a uniform electric field of strength 5.0 × 103 N C−1. What is the force on the charge?

  • (A) 1.6 × 10−15 N
  • (B) 1.6 × 10−22 N
  • (C) 6.4 × 10−16 N
  • (D) 1.6 × 10−12 N
Show Answer

Answer: A – F = QE = 3.2 × 10−19 × 5.0 × 103 = 1.6 × 10−15 N, acting in the direction of the field because the charge is positive.

Q5. An electron is released from rest between two oppositely charged parallel plates. How does it move?

  • (A) Towards the negative plate, in the direction of the field
  • (B) Towards the positive plate, opposite to the direction of the field
  • (C) Parallel to the plates, at constant speed
  • (D) It stays at rest because the field is uniform
Show Answer

Answer: B – The force on a charge is F = QE. For an electron Q is negative, so the force is opposite to the field direction: the electron is attracted to the positive plate and accelerates towards it.

Q6. Which pair of units are equivalent units for electric field strength?

  • (A) N C−1 and V m−1
  • (B) N C−1 and J m−1
  • (C) V m−1 and C N−1
  • (D) J C−1 and V m−1
Show Answer

Answer: A – Move a charge Q between plates a distance d apart with p.d. V. The work done is W = QV, and also W = Fd, so Fd = QV, giving F/Q = V/d, i.e. E = V/d. So field strength can be measured in volts per metre, and 1 V m−1 = 1 N C−1.

Q7. For an ideal pair of oppositely charged parallel plates, the electric field strength just outside the plates (away from the edges) is approximately

  • (A) equal to the field strength between the plates
  • (B) twice the field strength between the plates
  • (C) radial, like the field of a point charge
  • (D) zero
Show Answer

Answer: D – For ideal infinite parallel plates the fields of the two plates cancel outside and add between them, so the field exists only between the plates. Real plates show small “fringing” fields near the edges only.

Q8. The p.d. between a pair of plates is 500 V. How much work is done by the field when an electron (charge magnitude 1.6 × 10−19 C) moves from one plate to the other?

  • (A) 8.0 × 10−17 J
  • (B) 3.2 × 10−22 J
  • (C) 8.0 × 10−14 J
  • (D) 3.1 × 1016 J
Show Answer

Answer: A – W = QV = 1.6 × 10−19 × 500 = 8.0 × 10−17 J. This is the kinetic energy the electron gains (an energy of 500 eV).

Q9. The p.d. between a pair of parallel plates is doubled and, at the same time, the plate separation is doubled. What happens to the field strength between the plates?

  • (A) It doubles
  • (B) It halves
  • (C) It stays the same
  • (D) It increases by a factor of four
Show Answer

Answer: C – E = V/d. Doubling both V and d leaves the ratio V/d unchanged, so E stays the same.

Q10. A p.d. of 1500 V is applied across parallel plates separated by 3.0 cm. What is the field strength between the plates?

  • (A) 45 V m−1
  • (B) 5.0 × 102 V m−1
  • (C) 5.0 × 103 V m−1
  • (D) 5.0 × 104 V m−1
Show Answer

Answer: D – E = V/d = 1500 / 0.030 = 50 000 = 5.0 × 104 V m−1.

Motion of Charges in a Uniform Field

Q11. An electron (charge 1.6 × 10−19 C, mass 9.11 × 10−31 kg) is in a uniform electric field of strength 1.0 × 104 N C−1. What is its acceleration?

  • (A) 1.8 × 1011 m s−2
  • (B) 1.8 × 1013 m s−2
  • (C) 1.8 × 1015 m s−2
  • (D) 1.8 × 1017 m s−2
Show Answer

Answer: C – F = QE = 1.6 × 10−19 × 1.0 × 104 = 1.6 × 10−15 N. Then a = F/m = 1.6 × 10−15 / 9.11 × 10−31 = 1.8 × 1015 m s−2. The acceleration is constant because the field (and so the force) is uniform.

Q12. An electron is accelerated from rest through a p.d. of 100 V. What is its final speed? (Electron mass 9.11 × 10−31 kg.)

  • (A) 5.9 × 106 m s−1
  • (B) 1.9 × 106 m s−1
  • (C) 5.9 × 105 m s−1
  • (D) 1.9 × 107 m s−1
Show Answer

Answer: A – ½mv2 = QV, so v = √(2QV/m) = √(2 × 1.6 × 10−19 × 100 / 9.11 × 10−31) = √(3.5 × 1013) = 5.9 × 106 m s−1.

Q13. A proton (charge 1.6 × 10−19 C, mass 1.67 × 10−27 kg) is accelerated from rest through a p.d. of 100 V. What is its final speed?

  • (A) 5.9 × 106 m s−1
  • (B) 1.4 × 105 m s−1
  • (C) 1.4 × 107 m s−1
  • (D) 5.9 × 104 m s−1
Show Answer

Answer: B – The proton gains the same kinetic energy as the electron in Q12 (100 eV = 1.6 × 10−17 J), but its mass is about 1836 times larger. v = √(2QV/m) = √(2 × 1.6 × 10−19 × 100 / 1.67 × 10−27) = 1.4 × 105 m s−1.

Q14. A charged particle is projected into a uniform electric field at right angles to the field lines. What path does it follow while it is in the field?

  • (A) A circular arc
  • (B) A straight line at 45° to the field
  • (C) A parabola
  • (D) A helix
Show Answer

Answer: C – The motion is exactly like a projectile in gravity: the velocity component perpendicular to the field stays constant, while the component along the field increases steadily under the constant force. Constant velocity in one direction plus constant acceleration at right angles gives a parabolic path.

Q15. A charged particle enters a uniform electric field at right angles to the field lines. While it is between the deflecting plates, which quantity remains constant?

  • (A) Its speed
  • (B) Its kinetic energy
  • (C) The component of its velocity along the field
  • (D) The component of its velocity perpendicular to the field
Show Answer

Answer: D – The force acts only along the field, so there is no force (and no acceleration) perpendicular to the field. That component of velocity is unchanged, just as the horizontal velocity of a projectile is unchanged.

Q16. What kinetic energy does an electron gain when it is accelerated from rest through a p.d. of 250 V?

  • (A) 4.0 × 10−17 J
  • (B) 1.6 × 10−19 J
  • (C) 4.0 × 10−15 J
  • (D) 6.4 × 10−17 J
Show Answer

Answer: A – KE gained = QV = 1.6 × 10−19 × 250 = 4.0 × 10−17 J. This is an energy of 250 eV, since 1 eV = 1.6 × 10−19 J.

Q17. An electron and a proton enter the same uniform electric field with the same speed, at right angles to the field. Which statement about their deflections is correct?

  • (A) The proton is deflected much less, because the forces on the two particles are equal in size but the proton has a much larger mass
  • (B) The proton is deflected more, because it is positively charged
  • (C) Both particles are deflected by exactly the same amount
  • (D) The proton is not deflected at all
Show Answer

Answer: A – Both have charge of magnitude e, so both feel a force of the same size, F = eE (in opposite directions). But a = F/m, and the proton’s mass is about 1836 times the electron’s, so its acceleration, and hence its deflection in the same time, is far smaller.

Q18. What is the charge-to-mass ratio (specific charge) of the electron? (e = 1.6 × 10−19 C, me = 9.11 × 10−31 kg.)

  • (A) 1.8 × 108 C kg−1
  • (B) 1.8 × 1011 C kg−1
  • (C) 9.6 × 107 C kg−1
  • (D) 5.7 × 10−12 C kg−1
Show Answer

Answer: B – Q/m = 1.6 × 10−19 / 9.11 × 10−31 = 1.8 × 1011 C kg−1. This very large specific charge is why electrons are deflected so easily by electric (and magnetic) fields. Option C is the proton’s specific charge.

Q19. An electron travels horizontally at 2.0 × 106 m s−1 between deflecting plates of length 2.0 cm. The field between the plates is 2.0 × 103 V m−1. What is the electron’s deflection as it leaves the plates?

  • (A) 3.5 × 10−2 m
  • (B) 1.8 × 10−2 m
  • (C) 8.8 × 10−3 m
  • (D) 1.8 × 10−4 m
Show Answer

Answer: B – Time between the plates: t = 0.020 / 2.0 × 106 = 1.0 × 10−8 s. Acceleration: a = eE/me = 1.6 × 10−19 × 2.0 × 103 / 9.11 × 10−31 = 3.5 × 1014 m s−2. Deflection: y = ½at2 = 0.5 × 3.5 × 1014 × (1.0 × 10−8)2 = 1.8 × 10−2 m. Option A forgets the ½ in ½at2.

Q20. An alpha particle has charge +3.2 × 10−19 C and mass 6.6 × 10−27 kg. What is its acceleration in a uniform field of strength 2.0 × 104 N C−1?

  • (A) 4.8 × 1011 m s−2
  • (B) 1.8 × 1015 m s−2
  • (C) 9.7 × 108 m s−2
  • (D) 9.7 × 1011 m s−2
Show Answer

Answer: D – a = QE/m = 3.2 × 10−19 × 2.0 × 104 / 6.6 × 10−27 = 6.4 × 10−15 / 6.6 × 10−27 = 9.7 × 1011 m s−2. The double charge does not compensate for the much larger mass, so this is far below the electron’s acceleration in Q11. Option A uses only one electron charge.


Applications

Q21. In a cathode-ray oscilloscope, an electron beam passes between a pair of Y-plates and strikes a screen. The vertical deflection of the spot on the screen is

  • (A) proportional to the p.d. applied between the Y-plates
  • (B) proportional to the square of the p.d. between the Y-plates
  • (C) independent of the p.d. between the Y-plates
  • (D) inversely proportional to the p.d. between the Y-plates
Show Answer

Answer: A – E = V/d, so the deflecting force F = eE is proportional to V, and hence so are the acceleration and the deflection. This is why an oscilloscope can display a p.d. as a deflection: the vertical scale is calibrated in volts per division.

Q22. In an inkjet printer, each ink droplet is given an electric charge and then passes between a pair of deflecting plates held at a fixed p.d. The amount by which a droplet is deflected onto the paper is controlled by

  • (A) the speed of the droplet only
  • (B) the colour of the ink
  • (C) the size of the droplet only
  • (D) the electric charge put on the droplet
Show Answer

Answer: D – The printer electronics charge each droplet by a controlled amount. In the fixed field, F = QE, so a larger charge gives a larger deflection and the droplet lands higher on the page. Droplets given no charge fly straight on into a gutter and are recycled.

Q23. In Millikan’s oil-drop experiment, a droplet of mass 3.2 × 10−15 kg is held stationary by an upward electric field of strength 2.0 × 105 V m−1. What is the charge on the droplet? (g = 9.81 m s−2.)

  • (A) 3.2 × 10−19 C
  • (B) 8.0 × 10−20 C
  • (C) 1.6 × 10−19 C
  • (D) 1.6 × 10−18 C
Show Answer

Answer: C – For the droplet to be stationary, the electric force balances its weight: QE = mg, so Q = mg/E = 3.2 × 10−15 × 9.81 / 2.0 × 105 = 1.6 × 10−19 C – the charge of a single electron.

Q24. An oil droplet of mass 4.8 × 10−15 kg carries a charge of 3.2 × 10−19 C. What field strength is needed to hold it stationary? (g = 9.81 m s−2.)

  • (A) 1.5 × 105 V m−1
  • (B) 1.5 × 104 V m−1
  • (C) 3.0 × 105 V m−1
  • (D) 6.8 × 10−6 V m−1
Show Answer

Answer: A – QE = mg, so E = mg/Q = 4.8 × 10−15 × 9.81 / 3.2 × 10−19 = 1.5 × 105 V m−1.

Q25. A charged particle moves through a uniform electric field travelling exactly parallel to the field lines. Which statement is correct?

  • (A) It is speeded up or slowed down by the field, but it is not deflected sideways
  • (B) It experiences no force, because it moves along the field
  • (C) It is deflected into a circular path
  • (D) Its velocity remains exactly constant
Show Answer

Answer: A – The force F = QE acts along the field lines, so it can only change the particle’s speed along that line (speeding it up if a positive charge moves along E, slowing it down if it moves against E). There is no sideways component of force, so there is no sideways deflection.

Q26. A proton is fired into a uniform electric field in the direction exactly opposite to the field lines. What happens to the proton? (Ignore gravity.)

  • (A) It curves away sideways out of the field
  • (B) It slows down, stops momentarily, and then accelerates back along the same straight line
  • (C) It continues at constant speed in a straight line
  • (D) It spirals around the field lines
Show Answer

Answer: B – The force on the positive proton acts along the field, opposite to its motion, so it decelerates steadily to rest and then accelerates back the way it came – like a ball thrown vertically upwards in a gravitational field. Its path stays on the same straight line.

Q27. In a cathode-ray oscilloscope, the deflection produced by a given p.d. on the Y-plates can be increased by

  • (A) increasing the accelerating p.d. of the electron beam
  • (B) increasing the separation of the Y-plates
  • (C) decreasing the separation of the Y-plates
  • (D) shortening the Y-plates
Show Answer

Answer: C – E = V/d, so a smaller plate separation gives a stronger field and a larger deflecting force for the same applied p.d. Increasing the accelerating p.d. would do the opposite: a faster beam spends less time between the plates and is deflected less.

Q28. An oil droplet is held stationary in Millikan’s apparatus. The droplet then gains one extra electron. To hold it stationary again, the electric field strength must be

  • (A) increased
  • (B) decreased
  • (C) reversed in direction
  • (D) kept exactly the same
Show Answer

Answer: B – Balance requires E = mg/Q. The mass is unchanged but Q has increased, so a smaller field now provides the same upward force QE = mg. (Left unbalanced, the droplet would accelerate upwards because QE would exceed mg.)

Q29. Millikan measured the charges on many oil droplets and found they were always close to a whole-number multiple of 1.6 × 10−19 C. This was evidence that

  • (A) electrons have negligible mass
  • (B) oil droplets are always negatively charged
  • (C) electric fields between plates are uniform
  • (D) electric charge is quantised – it comes in units of the electron charge
Show Answer

Answer: D – Every measured charge was n × 1.6 × 10−19 C, where n is a whole number – the number of extra (or missing) electrons on the droplet. Charge never appeared in fractions of e, showing that charge is quantised.

Q30. A pair of plates is connected to a 600 V supply and the plates are 6.0 mm apart. An ink droplet between the plates carries a charge of 4.8 × 10−19 C. What is the force on the droplet?

  • (A) 4.8 × 10−17 N
  • (B) 2.9 × 10−16 N
  • (C) 4.8 × 10−14 N
  • (D) 8.0 × 10−17 N
Show Answer

Answer: C – First find the field: E = V/d = 600 / 0.0060 = 1.0 × 105 V m−1. Then F = QE = 4.8 × 10−19 × 1.0 × 105 = 4.8 × 10−14 N. Option B is QV, which is an energy, not a force.

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