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The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Current, Resistance, and Electromotive Force – Lecture 7 (University Physics II)

Overview

This lecture finishes the study of capacitors by treating dielectrics quantitatively, then opens a new chapter on current, resistance, and electromotive force. On the capacitor side it shows how a polarizable material inserted between the plates weakens the internal field, raises the capacitance, and lets us reinterpret the constant ε0\varepsilon_0 as the permittivity of empty space. It then explains what electric current actually is at the level of drifting electrons, defines resistivity and resistance, states Ohm's law, and introduces the emf and internal resistance of a real battery — closing with electrical power.

Key Concepts

Dielectrics and the dielectric constant

E=E0K,K≥1\begin{aligned} E &= \frac{E_0}{K}, \qquad K \ge 1 \end{aligned}

Permittivity

C=Kε0Ad=εAd,u=12Kε0E2=12εE2\begin{aligned} C &= \frac{K \varepsilon_0 A}{d} = \frac{\varepsilon A}{d}, \qquad u = \tfrac{1}{2}K\varepsilon_0 E^2 = \tfrac{1}{2}\varepsilon E^2 \end{aligned}

Dielectric strength vs. dielectric constant

Gauss's law with a dielectric

K∮E dA=Qfreeε0\begin{aligned} K \oint E \, dA &= \frac{Q_{\text{free}}}{\varepsilon_0} \end{aligned}

Fixed-charge vs. fixed-voltage rule. When a charged capacitor is disconnected from its source, the charge is fixed; when it stays connected to a source, the voltage is fixed. Everything else follows from C=Q/VC = Q/V.

Current and drift velocity

Resistivity, conductivity, and resistance

Electromotive force and real batteries

Electrical power

Worked Example

A 12 V battery with internal resistance r=2 Ωr = 2\ \Omega drives a single external resistor R=4 ΩR = 4\ \Omega. Find the current, the terminal voltage across the battery, and the power dissipated in RR.

Apply the loop rule E=I(r+R)\mathcal{E} = I(r + R), then the terminal-voltage and power relations:

I=Er+R=12 V2 Ω+4 Ω=2 AVab=E−Ir=12 V−(2 A)(2 Ω)=8 VPR=I2R=(2 A)2(4 Ω)=16 W\begin{aligned} I &= \frac{\mathcal{E}}{r + R} = \frac{12\ \text{V}}{2\ \Omega + 4\ \Omega} = 2\ \text{A} \\ V_{ab} &= \mathcal{E} - I r = 12\ \text{V} - (2\ \text{A})(2\ \Omega) = 8\ \text{V} \\ P_R &= I^2 R = (2\ \text{A})^2 (4\ \Omega) = 16\ \text{W} \end{aligned}

The battery's chemistry supplies EI=24 W\mathcal{E} I = 24\ \text{W}; of this, I2r=8 WI^2 r = 8\ \text{W} is wasted heating the battery internally, leaving the 16 W16\ \text{W} delivered to and dissipated by RR.

Summary