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PU-102 / L13 · lecture

Conductors and Capacitance

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with Gauss's law for electric fields and the definition of electric potential difference. Familiarity with basic circuit elements (resistors, capacitors) and exponential decay functions is also assumed.

Why this matters

Understanding how conductors behave in electrostatic equilibrium is the foundation for defining capacitance and analysing real-world energy storage devices.

Before we can discuss how capacitors store energy or how circuits charge and discharge, we must establish the rigorous boundary conditions that govern the interface between a conductor and the vacuum (or dielectric) surrounding it. This lecture bridges the gap between the microscopic motion of charges and the macroscopic circuit parameters used in engineering. By deriving the field structure at a conductor surface, we unlock the definition of capacitance for arbitrary geometries, which is essential for designing everything from microchips to high-voltage transmission lines.

The Electrostatic Shield

Imagine a metal sphere placed in an external electric field. If you were a test charge inside the metal, you would feel no force at all. This is not a coincidence; it is a requirement of electrostatic equilibrium. In a conductor, charges are free to move. If there were any electric field inside the material, these free charges would experience a force and begin to drift. They would continue moving until they rearranged themselves in such a way that the internal field they produce exactly cancels the external field. Once the charges stop moving, the system is in equilibrium. Consequently, the entire volume of the conductor becomes an equipotential region. The electric field lines, which always point from high to low potential, cannot penetrate this region; they must terminate perpendicularly on the surface. This perpendicularity is crucial because it ensures that the surface charge density is the sole source of the field just outside the conductor.

Boundary Conditions at the Surface

To quantify the field at the surface, we apply Gauss's law to a small pillbox that straddles the conductor boundary. Let the pillbox have area \(A\) and negligible thickness. Since the field inside the conductor is zero, the only flux contribution comes from the face outside the conductor. The flux through the side walls is zero because the field is normal to the surface. The flux through the outer face is \(E_{\perp} A\). The charge enclosed by the pillbox is the surface charge density \(\sigma\) times the area \(A\). Applying Gauss's law, \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\), we find that \(E_{\perp} A = \frac{\sigma A}{\varepsilon_0}\). This leads directly to the fundamental boundary condition for the electric field just outside a conductor in vacuum.

\[ \vec{E}_{\text{out}} = \frac{\sigma}{\varepsilon_0} \hat{n} \]

Here, \(\hat{n}\) is the unit normal vector pointing outward from the conductor. This result implies that the electric field strength is directly proportional to the local surface charge density. Regions of high curvature, such as the tip of a needle, will have a higher \(\sigma\) and thus a stronger local electric field. This is the physical basis of the lightning-rod (corona) effect: the field is strongest near sharp points, so the air there breaks down first.

Defining Capacitance

Capacitance is a geometric property of a system of conductors. It is defined as the ratio of the magnitude of the charge \(Q\) on one conductor to the potential difference \(V\) between two conductors. Mathematically, \(C = Q/V\). Because the potential difference is the line integral of the electric field, and the electric field is determined by the charge distribution via Gauss's law, capacitance depends only on the geometry and the permittivity of the medium, not on the amount of charge stored. For a parallel-plate capacitor with plate area \(A\) and separation \(d\), assuming \(d \ll \sqrt{A}\), the field is uniform and given by \(E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)\). The potential difference is \(V = Ed = Qd/(\varepsilon_0 A)\). Substituting this into the definition of capacitance yields the familiar formula.

\[ C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d} \]

This linear relationship between \(Q\) and \(V\) is a hallmark of linear dielectrics and ideal conductors. It allows us to treat the capacitor as a simple circuit element with a constant parameter \(C\), regardless of the operating voltage (within the breakdown limit of the dielectric).

Transient Response in RC Circuits

When a capacitor is connected to a voltage source through a resistor, the charge does not appear instantaneously. The rate of charging is limited by the resistance, which opposes the flow of current. Applying Kirchhoff's voltage law to a series circuit with emf \(\mathcal{E}\), resistance \(R\), and capacitance \(C\), we obtain the differential equation \( \mathcal{E} - IR - V_C = 0 \). Since \(I = dQ/dt\) and \(V_C = Q/C\), this becomes \( \mathcal{E} - R \frac{dQ}{dt} - \frac{Q}{C} = 0 \). Solving this first-order linear differential equation with the initial condition \(Q(0)=0\) gives the exponential charging curve.

\[ Q(t) = C\mathcal{E}\left(1 - e^{-t/\tau}\right) \]

The time constant \(\tau = RC\) characterizes the speed of the transient process. After one time constant, the capacitor is charged to approximately 63% of its final value. The current, which is the derivative of the charge, starts at its maximum value \(\mathcal{E}/R\) and decays exponentially to zero. This behavior is critical in timing circuits, filtering, and signal processing, where the precise control of charge accumulation and dissipation is required.

The derivations

◆ DERIVATION EMBED · D-046 · not re-derived here

Fields at Conductor Surfaces

In electrostatic equilibrium the macroscopic electric field inside a conductor vanishes, \(\vec{E}_{\text{in}}=\vec{0}\); the bulk and surface together form a single equipotential volume, \(V=\text{const}\); and just outside the surface the field is purely normal with magnitude fixed by the local surface charge density, \(\vec{E}_{\text{out}}=\dfrac{\sigma}{\varepsilon_0}\,\hat{n}\), where \(\hat{n}\) is the outward unit normal. As a corollary the volume charge density inside vanishes and all net charge lives on the surface.

\[ \vec{E}_{\text{in}}=\vec{0},\qquad \vec{E}_{\text{out}}=\frac{\sigma}{\varepsilon_0}\,\hat{n},\qquad V=\text{const on and in the conductor} \]
Open full derivation →
◆ DERIVATION EMBED · D-428 · not re-derived here

The RC Circuit: Charging, Discharging and the Time Constant

For a series circuit consisting of a battery of constant emf \(\mathcal{E}\), a resistor of resistance \(R\), and a capacitor of capacitance \(C\), the charge on the capacitor as a function of time, starting from a discharged state at \(t=0\), is \(Q(t) = C\mathcal{E}\left(1 - e^{-t/\tau}\right)\) where \(\tau = RC\); the current is \(I(t) = \frac{\mathcal{E}}{R}e^{-t/\tau}\) and the capacitor voltage is \(V_C(t) = \mathcal{E}\left(1 - e^{-t/\tau}\right)\). For a discharging capacitor with initial charge \(Q_0\), the charge decays as \(Q(t) = Q_0 e^{-t/\tau}\).

\[ Q(t) = C\mathcal{E}\left(1 - e^{-t/\tau}\right), \quad I(t) = \frac{\mathcal{E}}{R}e^{-t/\tau}, \quad V_C(t) = \mathcal{E}\left(1 - e^{-t/\tau}\right), \quad \tau = RC \]
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Tier 3 ยท watch out

A common misconception is that the electric field inside a conductor is zero because the charges are "stationary." In reality, the field is zero because any non-zero field would cause acceleration, violating the equilibrium condition. However, in a steady current scenario (non-electrostatic), a small electric field *does* exist inside the conductor to drive the drift of electrons. The boundary condition \(E_{\text{out}} = \sigma/\varepsilon_0\) strictly applies to electrostatic equilibrium where \(I=0\). C

Worked example

Problem. A parallel-plate capacitor with plate area \(A = 0.02\ \mathrm{m^2}\) and separation \(d = 1.0\ \mathrm{mm}\) is connected to a \(12\ \mathrm{V}\) battery through a resistor of \(R = 10\ \mathrm{k\Omega}\). Calculate the capacitance and the time required for the capacitor to charge to 90% of its final voltage.

  1. Calculate the capacitance using \(C = \varepsilon_0 A / d\). Using \(\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{F/m}\): \[ C = \frac{(8.85 \times 10^{-12}\ \mathrm{F/m})(0.02\ \mathrm{m^2})}{1.0 \times 10^{-3}\ \mathrm{m}} = 1.77 \times 10^{-10}\ \mathrm{F} \]
  2. Determine the time constant \(\tau = RC\): \[ \tau = (10 \times 10^3\ \Omega)(1.77 \times 10^{-10}\ \mathrm{F}) = 1.77 \times 10^{-6}\ \mathrm{s} \]
  3. Use the charging equation \(V_C(t) = \mathcal{E}(1 - e^{-t/\tau})\). We want \(V_C = 0.90 \mathcal{E}\): \[ 0.90 = 1 - e^{-t/\tau} \implies e^{-t/\tau} = 0.10 \] Taking the natural logarithm: \[ -\frac{t}{\tau} = \ln(0.10) \approx -2.303 \implies t = 2.303 \tau \]
  4. Calculate the final time: \[ t = 2.303 \times (1.77 \times 10^{-6}\ \mathrm{s}) \approx 4.08 \times 10^{-6}\ \mathrm{s} \]
\[ t \approx 4.1\ \mu\mathrm{s} \]

Units check. Resistance in Ohms \((\Omega)\) times Capacitance in Farads \((\mathrm{F} = \mathrm{C/V} = \mathrm{A \cdot s / V})\) yields \(\Omega \cdot \mathrm{A \cdot s / V}\). Since \(\Omega = \mathrm{V/A}\), the units become \((\mathrm{V/A}) \cdot (\mathrm{A \cdot s / V}) = \mathrm{s}\). The time constant has units of seconds, which is consistent.

CHECK YOURSELF

Three quick tests

  1. Why must the electric field be perpendicular to the surface of a conductor in electrostatic equilibrium?
    AnswerIf the field had a tangential component, it would exert a force on the surface charges, causing them to move. Equilibrium requires zero net force on free charges, so the tangential component must vanish.
  2. How does the capacitance of a parallel-plate capacitor change if the plate separation is doubled?
    AnswerIt is halved. Since \(C = \varepsilon_0 A / d\), doubling \(d\) divides the capacitance by 2.
  3. In an RC charging circuit, what is the current at \(t = \tau\)?
    AnswerThe current is \(I(\tau) = \frac{\mathcal{E}}{R} e^{-1} \approx 0.37 \frac{\mathcal{E}}{R}\). It has decayed to 37% of its initial value.
PROBLEMS

Problem set for PU-102 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-046 · D-428 · Threads: force, energy, fields