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AS & A-Level Physics 18 — Electric Fields and Potential

AS & A-Level Physics 18 — Electric Fields and Potential

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Independent Deckloop revision aligned with the Cambridge International AS & A Level Physics (9702) syllabus, 2025–2027. Not affiliated with or endorsed by Cambridge International Education. Chapter 18 of 26: Electric fields and potential. Concepts, worked applications and misconception checks.

Physics EN A-Level
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Force on a Charge in an Electric Field

Once the electric field strength EE at a point is known, the force FF experienced by any charge qq placed at that point can be calculated using the formula F=qEF = qE. This is a direct rearrangement of the definition of electric field strength. It's crucial to remember that FF and EE are vector quantities. If qq is positive, the force FF will be in the same direction as the electric field EE. If qq is negative, the force FF will be in the opposite direction to EE. The SI unit for force is Newtons (N), for charge is Coulombs (C), and for electric field strength is Newtons per Coulomb (N C−1^{-1}).

Key points

  • The force on a charge qq in an electric field EE is given by F=qEF = qE.
  • If qq is positive, FF is in the same direction as EE.
  • If qq is negative, FF is in the opposite direction to EE.
  • FF and EE are vector quantities.

Worked example

Question

An electron (charge −1.60×10−19-1.60 \times 10^{-19} C) is placed in a uniform electric field of strength 3.0×1033.0 \times 10^3 N C−1^{-1} directed vertically upwards. Calculate the magnitude and direction of the electric force on the electron.

Solution

1. Identify the given values: q=−1.60×10−19q = -1.60 \times 10^{-19} C, E=3.0×103E = 3.0 \times 10^3 N C−1^{-1} (upwards).
2. Use the formula F=qEF = qE.
3. Calculate the magnitude: F=∣(−1.60×10−19 C)×(3.0×103 N C−1)∣=4.8×10−16F = |(-1.60 \times 10^{-19} \text{ C}) \times (3.0 \times 10^3 \text{ N C}^{-1})| = 4.8 \times 10^{-16} N.
4. Determine the direction: Since the electron has a negative charge and the field is upwards, the force on the electron will be in the opposite direction to the field, i.e., downwards.

The magnitude of the force on the electron is 4.8×10−164.8 \times 10^{-16} N, directed vertically downwards.

Common pitfalls

  • Forgetting the vector nature: Simply calculating the magnitude without considering the direction for negative charges.
  • Mixing up units: Ensure all quantities are in SI units before calculation.

Prerequisites

  • Requires understanding the definition of electric field strength.
  • Understanding the sign of charge is crucial for force direction.
  • Force is a vector, requiring understanding of vector quantities.
Further resources