Questions: Parallel Plate Capacitor Geometry and Field

Question 1

Two parallel conducting plates carry surface charge densities +σ and −σ. Using the superposition principle, what is the electric field strength in the region outside the capacitor (beyond either plate)?

• A. σ/ε₀ — both plates contribute fields that add in the exterior region
• B. σ/(2ε₀) — only the nearer plate contributes outside the capacitor
• C. Zero — the fields from the two plates cancel exactly outside
• D. 2σ/ε₀ — the exterior field is stronger than the interior field

Answer: (C)

Each plate alone produces a field E = σ/(2ε₀) pointing away from itself on both sides. Outside the capacitor, the +σ plate pushes the field in one direction while the −σ plate pulls it in the opposite direction — they point antiparallel and cancel, giving zero net field. Between the plates, both contributions point in the same direction (from + to −) and add to give σ/ε₀. This cancellation in the exterior is the defining feature of the parallel plate geometry and follows directly from superposition.

Question 2

A parallel plate capacitor has capacitance C. If the plate separation is doubled while plate area and charge remain unchanged, what is the new capacitance?

• A. 2C — more separation means more room to store charge
• B. C√2 — separation enters the formula as a square root
• C. C/2 — capacitance is inversely proportional to plate separation
• D. C — separation only affects the field, not the capacitance

Answer: (C)

From C = ε₀A/d, doubling d halves C. Physically: with the same charge Q spread over the same area, the field E = σ/ε₀ is unchanged, but voltage V = Ed doubles (because the plates are farther apart). Since C = Q/V, doubling V at constant Q means C is halved. Option A reverses the relationship — larger separation requires more voltage to maintain the same charge, which means lower capacitance, not higher.

Question 3

The electric field between the plates of a parallel plate capacitor is exactly twice the field that either plate alone would produce at the same location.

• T. True
• F. False

Answer: True

Each plate, treated as an infinite sheet with surface charge density σ, produces a field of magnitude σ/(2ε₀) at any point. Between the plates, both contributions point in the same direction (from the positive plate toward the negative plate), so they add: E = σ/(2ε₀) + σ/(2ε₀) = σ/ε₀. This is precisely twice what a single sheet produces. Outside the plates, the contributions point in opposite directions and cancel to zero.

Question 4

Increasing the plate area of a parallel plate capacitor while keeping plate separation and total stored charge constant will increase the voltage across the capacitor.

• T. True
• F. False

Answer: False

Larger plate area at constant charge Q means the charge spreads out: surface charge density σ = Q/A decreases. The electric field E = σ/ε₀ therefore decreases, and the voltage V = Ed also decreases. Equivalently, C = ε₀A/d increases, and since V = Q/C, higher C at constant Q gives lower V. The voltage drops when area increases at constant charge — the intuition that 'bigger plates must mean more voltage' is wrong.

Question 5

Explain physically why the capacitance of a parallel plate capacitor decreases when the plate separation increases, assuming all else is equal.

Model answer: The surface charge density σ = Q/A is unchanged, so the electric field E = σ/ε₀ between the plates is unchanged. But the voltage across the capacitor is V = Ed — it increases linearly with separation d. Since capacitance is C = Q/V, and V grows while Q is fixed, C decreases as d increases. Physically, farther plates mean the same amount of stored charge requires a larger potential difference to maintain — the capacitor is less efficient at storing charge per unit voltage.

An alternative physical intuition: moving the plates farther apart weakens the influence of the charges on each other, so the mutual attraction that drives charge to the plates is reduced. To store the same charge with weaker attraction requires a larger applied voltage. Either way, the formula C = ε₀A/d encodes the inverse relationship cleanly.