Source: Discussion Question 5C
Tags: parallel-plate capacitor, capacitance formula, dielectric, surface charge density, electric field between plates, capacitor design, epsilon naught, P212, PHYS 212, UIUC
Difficulty: Intermediate Prerequisites: Gauss's Law, electric field of an infinite plane of charge, definition of capacitance (C = Q/ΔV).
This topic builds the parallel-plate capacitor from first principles. You start with two conducting plates, work out where the charges go, derive the electric field between them, and arrive at the capacitance formula C = ε₀A/d. Once you have that formula, you can design a capacitor to a specification, understand what happens when you change geometry or insert a dielectric, and predict how a capacitor responds to being moved apart while either isolated or connected to a battery. This is the physical foundation behind every "C" you plug into a circuit problem.
A parallel-plate capacitor stores charge on the inner faces of two conducting plates. The capacitance depends only on geometry and the material between the plates: C = ε₀A/d (or κε₀A/d with a dielectric). Changing the plate separation while the capacitor is isolated versus while it stays connected to a battery produces opposite effects on several quantities, and this is a favourite exam question.
Parallel-plate capacitor
Two large, flat conducting plates separated by an insulating gap. In simple terms, it is the simplest and most common model of a capacitor: two metal sheets with a gap between them.
Surface charge density (σ)
Charge per unit area on a surface, σ = Q/A, measured in C/m². Think of it as how tightly packed the charges are on the plate.
Permittivity of free space (ε₀)
The fundamental constant that sets the scale of electric interactions in vacuum: ε₀ = 8.854 × 10⁻¹² C²/(N·m²). It appears in Coulomb's law, Gauss's law, and the capacitance formula.
Dielectric constant (κ)
A dimensionless number (κ ≥ 1) describing how much an insulating material reduces the electric field compared to vacuum. Inserting a dielectric with κ = 2 doubles the capacitance. Think of it as the material's "capacitance multiplier."
Dielectric
An insulating material placed between the plates of a capacitor. Common examples: glass, paper, ceramic, plastic film.
When charge +Q is placed on one plate and −Q on the other:
All the charge migrates to the inner surfaces (the faces that look at each other)
The outer surfaces carry zero charge
This happens because the electric field inside a conductor must be zero. If there were charge on the outer surfaces, there would be a net field inside the metal, which is not allowed in electrostatic equilibrium.
Surface charge densities:
Inner surfaces: σᵢ⁺ = +Q/A and σᵢ⁻ = −Q/A
Outer surfaces: σₒ⁺ = 0 and σₒ⁻ = 0
Outside the plates (left of both, or right of both):
E = 0. The fields from the two plates cancel exactly.
Between the plates:
Each infinite sheet of charge produces a uniform field σ/(2ε₀). The two sheets' fields point in the same direction between the plates (both point from + toward −), so they add:
E = σ/ε₀ = Q/(ε₀A)
This field is uniform, meaning it has the same magnitude and direction everywhere between the plates (as long as you are far from the edges).
Step 1: The potential difference between the plates is:
ΔV = E · d = Qd/(ε₀A)
where d is the plate separation.
Step 2: Apply the definition of capacitance:
C = Q/ΔV = Q / [Qd/(ε₀A)] = ε₀A/d
The charge Q cancels out entirely, confirming that capacitance is a property of the geometry and material, not of the voltage or charge applied.
Plate area A: larger plates mean more room for charge, so C increases
Plate separation d: farther apart means weaker coupling, so C decreases
Material between the plates: inserting a dielectric with constant κ gives C = κε₀A/d
Capacitance does not depend on the voltage applied or the charge stored.
To build a capacitor with a target capacitance C_target:
Choose a plate area and separation that satisfy A/d = C_target/ε₀
Example: for C = 100 pF with air between the plates, you need A/d = 100 × 10⁻¹²/(8.854 × 10⁻¹²) ≈ 11.3 m. One possibility: A = 0.113 m² (roughly a 34 cm × 34 cm plate) with d = 1 cm.
If a client demands twice the capacitance from the same physical plates (same A, same d):
Insert a dielectric material with κ = 2 between the plates
This is the only option when the geometry is fixed
Suppose the capacitor is charged to some Q and then disconnected from the battery. The charge is trapped.
If you increase the separation from d to D:
Charge density σ: stays the same (charge is fixed, area is fixed)
Electric field E = σ/ε₀: stays the same (field depends on σ, not on d)
Potential difference ΔV = Ed: increases (same E, larger gap)
Capacitance C = ε₀A/d: decreases (d got bigger)
Now suppose the capacitor remains connected to a battery of voltage V₀ while the plates are moved apart. The battery enforces a fixed voltage.
If you increase the separation from d to D:
Capacitance C = ε₀A/D: decreases
Potential difference ΔV: stays the same (battery fixes it at V₀)
Charge Q = CV₀: decreases (C dropped, V is fixed, so charge flows back to the battery)
Charge density σ = Q/A: decreases
Electric field E = V₀/D: decreases (same voltage across a larger gap)
Parallel-plate capacitance:
C = ε₀A/d (vacuum or air)
C = κε₀A/d (with dielectric)
Electric field between plates:
E = σ/ε₀ = Q/(ε₀A) = ΔV/d
Potential difference:
ΔV = Ed = Qd/(ε₀A)
Surface charge density:
σ = Q/A
Key constant:
ε₀ = 8.854 × 10⁻¹² F/m
Every capacitor in electronics, from the tiny ceramic caps on a circuit board to the large electrolytic caps in a power supply, is governed by these principles. The reason capacitors in your phone are so small yet have useful capacitance is that manufacturers use ultra-thin dielectric layers (small d) with high-κ materials (large κ), and they stack many layers to increase effective area. The "supercapacitors" used in regenerative braking exploit the same C = κε₀A/d relationship with nanoscale gaps and enormous effective surface areas.
Students often think capacitance depends on the voltage applied. It does not. C = ε₀A/d is set by geometry and material. Changing V changes Q, not C.
Confusing the isolated case with the battery-connected case. When the capacitor is isolated, Q is fixed and V changes with plate separation. When connected to a battery, V is fixed and Q changes. These two scenarios are distinct and give opposite answers for several quantities. If the problem does not state which case applies, you cannot solve it.
Assuming the electric field between the plates depends on the separation. For a parallel-plate capacitor, E = σ/ε₀, which has no d in it. The field depends on charge density, not on how far apart the plates are (provided the capacitor is isolated). When connected to a battery, E does change with d, but only because σ changes.
Forgetting that charge resides only on the inner surfaces. The outer surfaces are charge-free in the ideal two-plate setup (no external fields).
⚠️ The "pull the plates apart" question (isolated vs. battery-connected) is one of the most common exam questions on this topic. Know both cases cold.
⚠️ Be able to derive C = ε₀A/d from scratch: field between plates, then ΔV = Ed, then C = Q/ΔV. This derivation is sometimes asked as a show-your-work problem.
⚠️ Dielectric insertion problems: know that inserting a dielectric increases C by a factor of κ, and be ready to work out what happens to Q, V, E, and U depending on whether the battery is connected or not.
⚠️ Units: ε₀ has units of F/m (farads per metre). If your answer for capacitance does not come out in farads, check your unit conversions, especially cm to m.
True or False: Doubling the plate area doubles the capacitance.
Fill in the blank: The electric field outside both plates of an ideal parallel-plate capacitor is ______.
True or False: If you double the plate separation of an isolated charged capacitor, the electric field between the plates doubles.
Fill in the blank: Inserting a dielectric with κ = 3 into a capacitor multiplies its capacitance by ______.
True or False: When a charged capacitor remains connected to a battery and the plates are pulled apart, the charge on the plates increases.
Answers: 1. True. 2. Zero. 3. False (it stays the same; E = σ/ε₀ and σ has not changed). 4. 3. 5. False (it decreases, because C decreases while V stays fixed).
Q: A parallel-plate capacitor has plates of area 0.5 m² separated by 2 mm of air. What is its capacitance?
A: C = ε₀A/d = (8.854 × 10⁻¹²)(0.5)/(0.002) = 2.21 × 10⁻⁹ F = 2.21 nF.
Q: The capacitor above is charged to 100 V and then disconnected from the battery. The plate separation is then increased to 5 mm. What is the new voltage across the plates?
A: Charge is fixed. Original Q = CV = 2.21 × 10⁻⁹ × 100 = 2.21 × 10⁻⁷ C. New C = ε₀A/D = (8.854 × 10⁻¹²)(0.5)/(0.005) = 8.854 × 10⁻¹⁰ F. New V = Q/C_new = 2.21 × 10⁻⁷ / 8.854 × 10⁻¹⁰ = 250 V.
Q: Instead, suppose the battery stays connected at 100 V while the plates are moved to 5 mm apart. What happens to the charge?
A: New C = 8.854 × 10⁻¹⁰ F. New Q = C_new × V = 8.854 × 10⁻¹⁰ × 100 = 8.854 × 10⁻⁸ C = 88.5 nC. The charge decreased (from 221 nC to 88.5 nC) because C decreased while V stayed fixed.
Q: Why does all the charge sit on the inner surfaces of the plates?
A: The electric field inside a conductor in electrostatic equilibrium must be zero. The only arrangement that produces zero net field inside both plates is for all the charge to reside on the inner faces, where the two sheets' external fields cancel in the conductor interior.
This connects directly to Gauss's Law (used to derive the electric field of an infinite sheet of charge, which is the starting point for the capacitance derivation). It also connects to energy storage in electric fields, since U = ½CV² and the energy density u = ½ε₀E² both follow from the parallel-plate model. Later, when you study dielectrics in detail, you will revisit why κ reduces the field and how polarisation of the dielectric material is responsible.
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