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9702 · 18.1

Electric fields and field lines — practice questions

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

Worked example 1

A point charge of +5.0 nC is placed in a vacuum. Calculate the electric field strength at a distance of 10 cm from the charge. (ϵ0=8.85×10−12 F m−1\epsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1})

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  1. Identify given values: Charge Q=+5.0×10−9Q = +5.0 \times 10^{-9} C, distance r=10 cm=0.10 mr = 10 \text{ cm} = 0.10 \text{ m}, permittivity of free space ϵ0=8.85×10−12 F m−1\epsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}.
  2. Recall the formula for electric field strength due to a point charge: E=14πϵ0Qr2E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}.
  3. Substitute the values into the formula: E=14π(8.85×10−12)5.0×10−9(0.10)2E = \frac{1}{4\pi (8.85 \times 10^{-12})} \frac{5.0 \times 10^{-9}}{(0.10)^2}.
  4. Calculate the value: The term 14πϵ0\frac{1}{4\pi\epsilon_0} is approximately 8.99×109 N m2C−28.99 \times 10^9 \text{ N m}^2 \text{C}^{-2}. So, E=(8.99×109)×5.0×10−90.01=4495 N C−1E = (8.99 \times 10^9) \times \frac{5.0 \times 10^{-9}}{0.01} = 4495 \text{ N C}^{-1}.
  5. State the final answer with appropriate significant figures and units: E≈4.5×103 N C−1E \approx 4.5 \times 10^3 \text{ N C}^{-1}. The direction is radially outwards from the positive charge.

Worked example 2

Two parallel metal plates are separated by 2.0 cm in a vacuum. A potential difference of 500 V is applied across them, creating a uniform electric field. An electron is released from rest at the surface of the negative plate. Calculate: (a) the electric field strength between the plates, (b) the force on the electron, and (c) the acceleration of the electron. (Use e=1.60×10−19e = 1.60 \times 10^{-19} C, me=9.11×10−31m_e = 9.11 \times 10^{-31} kg)

Show solution outline

This problem combines concepts of uniform fields and mechanics.

(a) Calculate Electric Field Strength (E)

  1. Identify given values: Potential difference V=500V = 500 V, distance d=2.0 cm=0.020 md = 2.0 \text{ cm} = 0.020 \text{ m}.
  2. Use the formula for a uniform field: E=V/dE = V/d.
  3. Substitute values: E=500 V0.020 m=25000 V m−1E = \frac{500 \text{ V}}{0.020 \text{ m}} = 25000 \text{ V m}^{-1}.
  4. The electric field strength is 2.5×104 N C−12.5 \times 10^4 \text{ N C}^{-1}, directed from the positive to the negative plate.

(b) Calculate the Force (F) on the electron

  1. Use the formula F=EqF = Eq. The charge qq is the elementary charge, ee.
  2. Substitute values: F=(25000 N C−1)×(1.60×10−19 C)F = (25000 \text{ N C}^{-1}) \times (1.60 \times 10^{-19} \text{ C}).
  3. Calculate the force: F=4.0×10−15 NF = 4.0 \times 10^{-15} \text{ N}.
  4. Since the electron is negatively charged, the force on it is opposite to the direction of the electric field (i.e., towards the positive plate).

(c) Calculate the Acceleration (a) of the electron

  1. Use Newton's Second Law: F=maF = ma.
  2. Rearrange for acceleration: a=F/ma = F/m.
  3. Substitute the force from part (b) and the mass of the electron: a=4.0×10−15 N9.11×10−31 kga = \frac{4.0 \times 10^{-15} \text{ N}}{9.11 \times 10^{-31} \text{ kg}}.
  4. Calculate the acceleration: a≈4.39×1015 m s−2a \approx 4.39 \times 10^{15} \text{ m s}^{-2}.
  5. The acceleration is in the same direction as the force, towards the positive plate.
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