Coulomb's-law and field problem clinic
≈ 48 minCoulomb's-law and field problem clinic
An electrostatics calculation is reliable when it separates the three jobs: magnitude, direction and meaning. Convert microcoulombs to coulombs, metres to SI distance and square the separation in Coulomb's law. Then use the signs to decide attraction or repulsion. For field questions, use E = F/q only after deciding which charge is the small test charge and whether the question asks field direction or force direction. A field is a property of a location created by source charges; it is not identical to the force on every possible charge placed there.
Work it through
At a point where E = 500 N C⁻¹ to the right, a +4.0 μC test charge experiences F = Eq = 500 × 4.0 × 10⁻⁶ = 2.0 × 10⁻³ N right. A −4.0 μC charge at the same point has a force of the same magnitude left. The field remains right in both cases by definition.
Mastery target
Calculate force from field or Coulomb's law, state direction using a test-charge convention and explain why a negative charge reverses force but not field direction.
At a point, an electric field points right. What happens to a negative test charge placed there?
Name the key physics term from Coulomb's-law and field problem clinic that best fits the explanation and visual model.
A field of 500 N C⁻¹ acts on a +4.0 μC test charge. What force magnitude in N acts on it?
Which statement corrects a common misunderstanding in Coulomb's-law and field problem clinic?