Difficulty: Intermediate | Prerequisites: Electric field from parallel plates, electric potential, basic circuit concepts (series vs parallel)
Big picture: Capacitors store energy in an electric field between conducting plates. They appear in nearly every electronic device and are the bridge between electrostatics (static charges and fields) and circuits (charges in motion). This topic covers how to identify series and parallel connections, how to compute equivalent capacitance, and how a dielectric slab changes the field and capacitance. Mastering this section gives you the foundation for RC circuits and AC circuits later in the course.
Capacitors in parallel share the same voltage and their capacitances add. Capacitors in series share the same charge and their reciprocals add. A dielectric inserted between the plates reduces the field by a factor of κ and increases the capacitance by a factor of κ. When only part of the gap is filled, treat the configuration as capacitors in series.
Capacitance (C)
The ratio of charge to voltage for a capacitor: C = Q/V, measured in farads (F). A larger capacitance means the device stores more charge per volt. In simple terms, capacitance measures how much charge a capacitor can hold for a given voltage across it.
Parallel plate capacitor
Two flat conducting plates of area A separated by distance d. The capacitance is C = ε₀A/d (with no dielectric). In simple terms, bigger plates and a smaller gap give more capacitance.
Series connection
Two circuit elements are in series if they carry the same current (or, for capacitors, the same charge). They share charge, not voltage. To find equivalent capacitance: 1/C_eq = 1/C₁ + 1/C₂ + ...
Parallel connection
Two circuit elements are in parallel if they have the same voltage across them. They share voltage, not charge. To find equivalent capacitance: C_eq = C₁ + C₂ + ...
Dielectric constant (κ)
A dimensionless number (κ ≥ 1) characterising an insulating material. Inserting a dielectric between the plates of a capacitor reduces the electric field by a factor of κ and increases the capacitance by a factor of κ. In simple terms, the dielectric is a polarisable material that partially cancels the field, allowing the capacitor to hold more charge at the same voltage.
Dielectric slab (partial fill)
When a dielectric fills only part of the gap between the plates, the capacitor behaves as two capacitors in series: one with the dielectric (thickness t, field E₀/κ) and one with vacuum (thickness d − t, field E₀).
Two capacitors are in parallel if both ends of one are connected directly to both ends of the other, so they share the same two nodes (and therefore the same voltage).
Two capacitors are in series if they are connected end-to-end with no branch in between, so the same charge must flow onto one plate and off the other.
In the circuit from the exam (C₁ = 6 μF, C₂ = 8 μF, C₃ = 12 μF, C₄ = 10 μF), C₃ and C₄ are connected in series. They are not in parallel because they do not share the same pair of nodes.
Using the circuit described (with C₁ and C₂ in parallel, then that combination in series with C₃, and C₄ in series with C₃):
First, recognise the topology from the circuit diagram. C₁ and C₂ are in parallel:
C₁₂ = C₁ + C₂ = 6 + 8 = 14 μF.
C₃ and C₄ are in series:
1/C₃₄ = 1/C₃ + 1/C₄ = 1/12 + 1/10 = (10 + 12)/120 = 22/120
C₃₄ = 120/22 ≈ 5.45 μF.
Then C₁₂ and C₃₄ are in series:
1/C₁₂₃₄ = 1/14 + 1/5.45 = (5.45 + 14)/(14 × 5.45) = 19.45/76.36
C₁₂₃₄ ≈ 76.36/19.45 ≈ 3.93 μF.
(Note: the exact answer depends on the circuit topology, which varies by exam version. With the given answer choices, work through the topology that yields one of the listed values. The answer C₁₂₃₄ = 6.1 μF corresponds to a different connection order where C₁₂ is in parallel with C₃₄, or another arrangement. Always trace the circuit carefully.)
If Q₁ = 5.83 μC and C₁ = 6 μF, then V₁ = Q₁/C₁ = 5.83/6 ≈ 0.97 V.
Since C₁ and C₂ are in parallel, V₂ = V₁ = 0.97 V.
The charge on C₂ is Q₂ = C₂ × V₂ = 8 × 0.97 = 7.76 μC.
If C₃ is in series with the parallel combination, it carries the total charge Q₁₂ = Q₁ + Q₂ = 5.83 + 7.76 = 13.59 μC.
But if C₃ and C₄ are in series with each other, they carry the same charge, and the analysis depends on the exact circuit topology. The voltage across C₃: V₃ = Q₃/C₃.
With the exam answer V₃ ≈ 1.1 V and C₃ = 12 μF, the charge on C₃ would be Q₃ = 12 × 1.1 = 13.2 μC.
The battery voltage is the sum of voltage drops across elements in series from one terminal to the other: V_battery ≈ 0.97 + 1.1 ≈ 2.1 V (depending on the series path from one battery terminal to the other).
A capacitor with plate area A = 5 cm² = 5 × 10⁻⁴ m², plate separation d = 1 mm = 10⁻³ m, and a dielectric slab (κ = 1.5) filling half the gap (thickness d/2).
The vacuum field: E₀ = σ/ε₀ = Q/(ε₀A).
With the charge fixed at Q = 6 × 10⁻¹¹ C:
E₀ = Q/(ε₀A) = (6 × 10⁻¹¹) / (8.85 × 10⁻¹² × 5 × 10⁻⁴) = (6 × 10⁻¹¹) / (4.425 × 10⁻¹⁵) ≈ 1.356 × 10⁴ V/m.
In the vacuum half (thickness d/2): E_vac = E₀ ≈ 1.356 × 10⁴ V/m.
In the dielectric half (thickness d/2): E_diel = E₀/κ = 1.356 × 10⁴ / 1.5 ≈ 9.04 × 10³ V/m.
Total potential difference: ΔV = E_vac × (d/2) + E_diel × (d/2) = (d/2)(E₀ + E₀/κ) = (d/2) × E₀ × (1 + 1/κ).
ΔV = (0.5 × 10⁻³)(1.356 × 10⁴)(1 + 1/1.5) = (0.5 × 10⁻³)(1.356 × 10⁴)(1.667) ≈ 11.3 V.
Looking at answer choices, ΔV ≈ 11 V.
Capacitance of the configuration:
C = Q/ΔV = (6 × 10⁻¹¹) / 11.3 ≈ 5.3 × 10⁻¹² F = 5.3 pF.
Alternatively, treat as two capacitors in series: C_vac = ε₀A/(d/2) and C_diel = κε₀A/(d/2).
1/C = 1/C_vac + 1/C_diel = (d/2)/(ε₀A) + (d/2)/(κε₀A) = (d/2ε₀A)(1 + 1/κ).
C = 2κε₀A / [d(κ + 1)] = 2(1.5)(8.85 × 10⁻¹²)(5 × 10⁻⁴) / [10⁻³(2.5)] ≈ 5.3 pF.
Parallel plate capacitor (no dielectric):
C = ε₀A / d
With a dielectric filling the entire gap:
C = κε₀A / d
Capacitors in series:
1/C_eq = 1/C₁ + 1/C₂ + ...
Capacitors in parallel:
C_eq = C₁ + C₂ + ...
Charge-voltage relationship:
Q = CV
Field in a dielectric (with the charge fixed):
E = E₀ / κ = Q / (κε₀A)
Potential difference with a partial dielectric (filling thickness t of a gap d):
ΔV = (Q / ε₀A) × [(d − t) + t/κ]
Energy stored in a capacitor:
U = ½CV² = ½Q²/C = ½QV
Capacitors with dielectrics are in every power supply, phone charger, and camera flash unit. The dielectric serves a dual purpose: it prevents the plates from shorting out (it is an insulator) and it increases the capacitance, allowing more energy storage in a smaller package.
Identifying series and parallel combinations is a core skill for circuit analysis. Real circuits contain dozens or hundreds of capacitors, and reducing them to an equivalent capacitance is the first step in understanding the circuit's behaviour.
Students often confuse which quantity is shared in series versus parallel. Series capacitors share charge (the same Q on each). Parallel capacitors share voltage (the same V across each). This is the opposite of resistors, which share current in series and voltage in parallel.
When a dielectric fills only part of the gap, students sometimes multiply the whole capacitance by κ. That would be correct only if the dielectric filled the entire gap. A partial fill requires the series-combination approach: one capacitor for the dielectric region and one for the vacuum region.
Students frequently forget to convert units: areas from cm² to m², distances from mm to m, charges from μC to C. One wrong power of ten makes the entire answer wrong.
A common error: assuming that inserting a dielectric always increases the voltage. If the charge is fixed (isolated capacitor, not connected to a battery), inserting a dielectric decreases the voltage (because C goes up and Q = CV, so V = Q/C goes down). If the capacitor stays connected to a battery (voltage fixed), inserting a dielectric increases the charge instead.
⚠️ "Are C₃ and C₄ in series or parallel?" is a topology question. Trace the circuit path: if the same charge must flow through both, they are in series. If they share the same pair of nodes (same voltage), they are in parallel. If neither condition is met, they are neither, and you cannot use the simple formulas directly.
⚠️ The voltage across individual capacitors is found by working from known quantities. If you know Q on one capacitor and its capacitance, V = Q/C gives its voltage. Use the sharing rules (series shares Q, parallel shares V) to propagate through the circuit.
⚠️ For the partial dielectric problem, the exam tests whether you can set up the series combination correctly. Draw the two regions, assign each its own capacitance, and combine in series.
⚠️ Always check whether the problem says "charge is fixed" (isolated capacitor) or "voltage is fixed" (connected to a battery). The answer is different in each case.
True or false: Two capacitors in series have the same voltage across them.
Fill in the blank: For capacitors in parallel, the equivalent capacitance is the ________ of the individual capacitances.
True or false: Inserting a dielectric into a capacitor always increases its capacitance.
Fill in the blank: The SI unit of capacitance is the ________.
True or false: When a dielectric fills only half the gap of a parallel plate capacitor, the configuration is equivalent to two capacitors in parallel.
Q: C₁ = 6 μF and C₂ = 8 μF are in parallel. What is their combined capacitance?
A: C₁₂ = C₁ + C₂ = 6 + 8 = 14 μF.
Q: If the charge on C₁ (6 μF) is 5.83 μC, what is the voltage across it?
A: V₁ = Q₁/C₁ = 5.83 μC / 6 μF ≈ 0.97 V.
Q: A parallel plate capacitor (A = 5 cm², d = 1 mm) has a dielectric slab (κ = 1.5) filling half the gap. What is the capacitance?
A: Treat as two capacitors in series. C = 2κε₀A / [d(κ + 1)] ≈ 5.3 pF.
Q: For the same capacitor with fixed charge Q = 6 × 10⁻¹¹ C, what is the potential difference between the plates?
A: ΔV = Q/C = (6 × 10⁻¹¹) / (5.3 × 10⁻¹²) ≈ 11 V. Alternatively, compute the field in each region, multiply by the thickness, and sum.
Q: What happens to the voltage across an isolated charged capacitor (not connected to a battery) when a dielectric is inserted?
A: The voltage decreases. The capacitance increases by a factor of κ, and since Q is fixed, V = Q/C decreases by the same factor.
Capacitance ties directly back to Gauss's law and the parallel plate field (E = σ/ε₀ between the plates). It also connects forward to energy storage (U = ½CV²), which appears in RC circuit problems where a capacitor charges and discharges through a resistor. The series/parallel reduction technique is the same one you will use for resistors in DC circuits, with the formulas swapped (series resistors add, parallel resistors use reciprocals, the opposite of capacitors).
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