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Topic 5 of 15

Capacitance & Dielectrics

A capacitor stores electrical energy by accumulating separated charge on two conductors. Capacitors are everywhere in electronics — from smoothing power supplies to tuning radio receivers to storing energy in defibrillators. The key insight is that the energy is stored in the electric field between the plates, not in the charges themselves.

Key Concepts

Capacitance

The capacitance

C

of a capacitor is the ratio of the charge stored on each plate to the potential difference between the plates:

C = Q/V

. Unit: farad (F = C/V). A larger capacitance stores more charge per volt.

Parallel-Plate Capacitor

Two large parallel conducting plates of area

A

separated by gap

d

. Capacitance:

C = \varepsilon_0 A/d

. Larger area and smaller gap both increase capacitance.

Dielectric Materials

Inserting an insulating material (dielectric) between the plates increases capacitance by the dielectric constant

\kappa > 1

C = \kappa\varepsilon_0 A/d

. The dielectric's polar molecules align with the field, reducing the effective electric field and allowing more charge to be stored at the same voltage.

Energy Stored in a Capacitor

The energy stored is

U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV

. This energy resides in the electric field between the plates.

Capacitors in Series

Capacitors in series all carry the same charge

Q

; their voltages add. Equivalent capacitance:

1/C_{\text{eq}} = 1/C_1 + 1/C_2 + \cdots

. The equivalent is always less than the smallest individual capacitor.

Capacitors in Parallel

Capacitors in parallel all share the same voltage

V

; their charges add. Equivalent capacitance:

C_{\text{eq}} = C_1 + C_2 + \cdots

. The equivalent is the sum.

Key Equations

Capacitance definition

C = \frac{Q}{V}

Charge Q stored per unit voltage V. Unit: farad (F).

Parallel-plate capacitor

C = \frac{\kappa\varepsilon_0 A}{d}

κ = dielectric constant (1 for air), ε₀ = 8.85×10⁻¹² F/m, A = plate area, d = plate separation.

Energy stored

U = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}

Energy (in joules) stored in the electric field of the capacitor.

Series combination

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Equivalent capacitance for capacitors connected in series (same Q, voltages add).

Parallel combination

C_{\text{eq}} = C_1 + C_2 + \cdots

Equivalent capacitance for capacitors connected in parallel (same V, charges add).

Worked Example

Parallel-Plate Capacitor with a Battery

Problem

A parallel-plate capacitor has plate area $A = 0.020$ m² and separation $d = 2.0$ mm (air gap, $\kappa = 1$ ). It is connected to a $12$ V battery. Find (a) the capacitance, (b) the charge stored, and (c) the energy stored.

Solution

(a) Capacitance:

C = \frac{\varepsilon_0 A}{d} = \frac{(8.85\times10^{-12})(0.020)}{2.0\times10^{-3}} = 88.5\text{ pF}

(b) Charge stored at $V = 12$ V:

Q = CV = (88.5\times10^{-12})(12) = 1.06\text{ nC}

U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(88.5\times10^{-12})(12)^2 = 6.37\text{ nJ}

Answer C = 88.5 pF; Q = 1.06 nC; U = 6.37 nJ.

Practice

Exercises

7 problems

1 of 7

A capacitor stores $Q = 60$ nC of charge when connected to a $V = 12$ V source. What is its capacitance (in nF)?

Your answer

2 of 7

A parallel-plate capacitor has plate area $A = 0.050$ m² and separation $d = 1.0$ mm with air between the plates. What is its capacitance (in pF)?

Your answer

3 of 7

How much energy (in mJ) is stored in a $C = 10\,\mu$ F capacitor charged to $V = 100$ V?

Your answer

4 of 7

Two capacitors $C_1 = 4\,\mu$ F and $C_2 = 6\,\mu$ F are connected in **series**. What is the equivalent capacitance (in $\mu$ F)?

Your answer

μF

5 of 7

The same two capacitors $C_1 = 4\,\mu$ F and $C_2 = 6\,\mu$ F are now connected in **parallel**. What is the equivalent capacitance (in $\mu$ F)?

Your answer

μF

6 of 7

A capacitor has capacitance $C_0 = 200$ pF with air between its plates. A dielectric with $\kappa = 3.5$ is inserted to fill the gap. What is the new capacitance (in pF)?

Your answer

7 of 7

A parallel-plate capacitor (air, $\kappa=1$ ) has plate area $A = 0.040$ m² and must have capacitance $C = 100$ pF. What plate separation $d$ (in mm) is needed?

Your answer

Key Takeaways

Capacitance measures how much charge is stored per volt — larger $C$ means more charge for the same voltage.
For a parallel plate: $C \propto A/d$ — bigger plates or smaller gap increases capacitance.
A dielectric multiplies $C$ by $\kappa$ because aligned molecules reduce the internal field, allowing more charge for the same $V$ .
In series: equivalent $C$ is less than the smallest (reciprocals add). In parallel: equivalent $C$ is the sum (capacitances add).
Energy stored scales as $V^2$ — doubling the voltage quadruples the stored energy.