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9702 · 18.2

Uniform electric fields — practice questions

Practice and worked examples for 9702 Uniform electric fields. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Two parallel plates are separated by 2.0 cm and have a potential difference of 500 V across them. a) Calculate the electric field strength between the plates. b) What force does an electron (charge = −1.6×10−19-1.6 \times 10^{-19} C) experience in this field?

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  1. Convert plate separation to metres: d=2.0 cm=0.02 md = 2.0 \text{ cm} = 0.02 \text{ m}.
  2. Calculate electric field strength: E=ΔVd=500 V0.02 m=25000 V m−1E = \frac{\Delta V}{d} = \frac{500 \text{ V}}{0.02 \text{ m}} = 25000 \text{ V m}^{-1} (or N C⁻¹).
  3. Calculate the magnitude of the force on the electron: F=∣Eq∣=(25000 N C−1)×(1.6×10−19 C)=4.0×10−15 NF = |Eq| = (25000 \text{ N C}^{-1}) \times (1.6 \times 10^{-19} \text{ C}) = 4.0 \times 10^{-15} \text{ N}.
  4. State the direction of the force: Since the electron is negative, the force is in the opposite direction to the electric field lines (i.e., towards the positive plate).

Worked example 2

An electron (mass 9.11×10−319.11 \times 10^{-31} kg, charge −1.6×10−19-1.6 \times 10^{-19} C) is fired horizontally with a speed of 2.0×1072.0 \times 10^7 m s⁻¹ into a uniform electric field. The field is created by two parallel plates, 5.0 cm long and 1.5 cm apart, with a potential difference of 300 V. Calculate the vertical deflection of the electron as it exits the plates.

Show solution outline
  1. Calculate Electric Field Strength (E): The field is uniform between the plates. E=ΔVd=300 V0.015 m=20000 V m−1E = \frac{\Delta V}{d} = \frac{300 \text{ V}}{0.015 \text{ m}} = 20000 \text{ V m}^{-1}.
  2. Calculate Vertical Force (F_y): The force on the electron is vertical (upwards, towards the positive plate). Fy=∣q∣E=(1.6×10−19 C)×(20000 N C−1)=3.2×10−15 NF_y = |q|E = (1.6 \times 10^{-19} \text{ C}) \times (20000 \text{ N C}^{-1}) = 3.2 \times 10^{-15} \text{ N}.
  3. Calculate Vertical Acceleration (a_y): Using Newton's second law, F=maF=ma. ay=Fyme=3.2×10−15 N9.11×10−31 kg=3.51×1015 m s−2a_y = \frac{F_y}{m_e} = \frac{3.2 \times 10^{-15} \text{ N}}{9.11 \times 10^{-31} \text{ kg}} = 3.51 \times 10^{15} \text{ m s}^{-2}.
  4. Calculate Time of Flight (t): The horizontal velocity is constant. The time spent between the plates depends on the plate length and horizontal speed. t=lengthhorizontal speed=0.050 m2.0×107 m s−1=2.5×10−9 st = \frac{\text{length}}{\text{horizontal speed}} = \frac{0.050 \text{ m}}{2.0 \times 10^7 \text{ m s}^{-1}} = 2.5 \times 10^{-9} \text{ s}.
  5. Calculate Vertical Deflection (s_y): Use the kinematic equation s=ut+12at2s = ut + \frac{1}{2}at^2. The initial vertical velocity (uyu_y) is zero. sy=(0)(t)+12ayt2=12×(3.51×1015 m s−2)×(2.5×10−9 s)2s_y = (0)(t) + \frac{1}{2}a_y t^2 = \frac{1}{2} \times (3.51 \times 10^{15} \text{ m s}^{-2}) \times (2.5 \times 10^{-9} \text{ s})^2 sy=12×(3.51×1015)×(6.25×10−18)=0.01097 ms_y = \frac{1}{2} \times (3.51 \times 10^{15}) \times (6.25 \times 10^{-18}) = 0.01097 \text{ m}.
  6. Final Answer: The vertical deflection is 0.0110.011 m or 1.11.1 cm. (Since this is less than half the plate separation of 1.5 cm, the electron exits without hitting the plate).
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