Two positive charges +q₁ and +q₂ are separated by 1 meter. A student computes the electric field at the midpoint by finding the magnitudes |E₁| = kq₁/(0.5)² and |E₂| = kq₂/(0.5)², then adds them as scalars. What is wrong with this approach?
AShe should use r = 1 m (the full separation), not 0.5 m
BShe treated the fields as scalars. At the midpoint, E₁ points away from q₁ (toward q₂) and E₂ points away from q₂ (toward q₁) — they point in opposite directions and must be added as vectors, potentially partially or fully canceling
CThe formula E = kq/r² applies only to negative charges; for positive charges a different formula is needed
DShe should average the fields, not add them, since there are two source charges
Electric field contributions from multiple charges must be added as vectors, not scalars. At the midpoint between two positive charges, E₁ points from q₁ toward the midpoint (i.e., in the +x direction if q₂ is to the right), and E₂ points from q₂ toward the midpoint (i.e., in the −x direction). They oppose each other. If q₁ = q₂, the fields cancel exactly and the net field is zero. Adding the magnitudes would give a nonzero answer regardless — the classic error of ignoring direction when applying superposition.
Question 2 Multiple Choice
Which statement best explains why the electric field framework is more than just a notational convenience for restating Coulomb's law?
AIt simplifies calculations by avoiding the need to track vector directions
BIt separates field creation (source charge) from field response (test charge), which becomes physically necessary when charges move — field changes propagate at the speed of light, so the interaction cannot be instantaneous
CIt eliminates the need for Coulomb's constant k in calculations involving multiple charges
DIt is purely a notational convenience; Coulomb's law and the field description are always physically equivalent
For static charges, Coulomb's law and the field description give identical results. But when charges are moving or accelerating, the field picture becomes physically essential: changes in a charge's position create changes in the field that propagate outward at c (the speed of light), not instantaneously. The 'action at a distance' implied by Coulomb's law breaks down. The field is a real physical entity with energy stored in it — not just a bookkeeping device — and this distinction is what leads ultimately to electromagnetic waves and special relativity.
Question 3 True / False
The electric field at a point in space exists primarily if there is a test charge placed there to detect it.
TTrue
FFalse
Answer: False
The electric field is a property of space itself, created by source charges and existing independently of whether any test charge is present to measure it. We *define* the field in terms of the force it would exert on a hypothetical positive test charge, but that test charge need not actually be there. This field-as-a-physical-entity picture is crucial: it is the field (not the source charges directly) that exerts forces on other charges, and field energy propagates through space even in the absence of charges.
Question 4 True / False
When computing the electric field at a point due to multiple charges, each charge's vector contribution uses a distance r measured from that specific source charge to the field point.
TTrue
FFalse
Answer: True
This is the most operationally important rule of superposition for electric fields. Each charge 'sees' the field point at its own distance and in its own direction. A common error is to compute |E| using the distance from the first charge and then reuse the same r for all other charges. Each contribution E_i = kq_i/r_i² r̂_i requires its own r_i (distance from charge i to the field point) and its own r̂_i (unit vector pointing from charge i toward the field point). Only after computing all individual vectors are they added component by component.
Question 5 Short Answer
Explain why the direction of the electric field is defined as the force on a positive test charge, and what happens to the field direction when the source charge is negative.
Think about your answer, then reveal below.
Model answer: The convention defines E as the force per unit charge that a small positive test charge would experience: E = F/q₀ where q₀ > 0. This means the field direction is 'the way a positive charge would be pushed.' For a positive source charge, the force on a positive test charge is repulsive — pointing away from the source — so E points radially outward. For a negative source charge, the force on a positive test charge is attractive — pointing toward the source — so E points radially inward. The field direction is always defined by the source's sign, regardless of what charge you place there.
Using a positive test charge as the reference is a convention chosen for consistency: the field direction and force direction are identical for positive test charges, and opposite for negative ones. This means field lines tell you exactly which way a positive charge would accelerate. If you place a negative charge q at a point where E points right, the force on it is −qE, which points left — the force and field directions are opposed for negative charges. The field itself hasn't changed; the sign of the responding charge determines whether the force is parallel or antiparallel to E.