where $Q$ is charge on each plate, $V$ is potential difference.

Units: 1 farad (F) = 1 coulomb/volt

Parallel Plate Capacitor

Area $A$, separation $d$:

$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$

$V = Ed = \frac{Qd}{\epsilon_0 A}$

$C = \frac{Q}{V} = \frac{\epsilon_0 A}{d}$

Other Geometries

Cylindrical Capacitor

Inner radius $a$, outer radius $b$, length $L$:

$C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}$

Spherical Capacitor

Inner radius $a$, outer radius $b$:

$C = 4\pi\epsilon_0\frac{ab}{b-a}$

Isolated Sphere

Radius $R$ (other conductor at infinity):

$C = 4\pi\epsilon_0 R$

Capacitors in Series

Same charge $Q$ on each:

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

Two capacitors: $C_{eq} = \frac{C_1C_2}{C_1 + C_2}$

Capacitors in Parallel

Same voltage $V$ across each:

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

Energy Stored in Capacitor

Work to charge capacitor:

$W = \int_0^Q V \, dq = \int_0^Q \frac{q}{C} \, dq = \frac{Q^2}{2C}$

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

Energy density (parallel plate): $u = \frac{U}{Ad} = \frac{1}{2}\epsilon_0 E^2$

Dielectrics

Insulating material inserted between plates:

Dielectric constant: $\kappa$ (or $K$)

With dielectric:

• Capacitance: $C = \kappa C_0$
• Electric field: $E = E_0/\kappa$
• Potential: $V = V_0/\kappa$ (if charge constant)

Permittivity of material: $\epsilon = \kappa\epsilon_0$

Modified equations: $C = \frac{\kappa\epsilon_0 A}{d}$

$u = \frac{1}{2}\kappa\epsilon_0 E^2$

Dielectric Breakdown

Maximum field before dielectric breaks down:

Air: ~$3 \times 10^6$ V/m

Different materials have different breakdown strengths.

Gauss's Law with Dielectrics

$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{free}}{\kappa\epsilon_0}$

or using electric displacement $\vec{D} = \kappa\epsilon_0\vec{E}$:

$\oint \vec{D} \cdot d\vec{A} = Q_{free}$

Polarization

Dielectric polarizes in electric field:

Polarization: $\vec{P} = \kappa\epsilon_0(\kappa - 1)\vec{E}$

Bound surface charge: $\sigma_b = P$

This reduces net field inside dielectric.

Energy with Dielectric

If dielectric inserted with:

Constant charge: $U_f = U_i/\kappa$ (energy decreases)

Constant voltage: $U_f = \kappa U_i$ (energy increases)

Force on dielectric:

Dielectric pulled into capacitor (lower energy state).

$C_{12} = 2.4 \text{ μF}$

C₁₂ in parallel with C₃: $C_{eq} = C_{12} + C_3 = 2.4 + 3.0$

$C_{eq} = \boxed{5.4 \text{ μF}}$

(b) Charges:

Total charge from battery: $Q_{total} = C_{eq}V = (5.4)(24) = 129.6 \text{ μC}$

For parallel combination:

• Q₃ = C₃V = (3.0)(24) = 72 μC
• Q₁₂ = C₁₂V = (2.4)(24) = 57.6 μC

In series, C₁ and C₂ have same charge:

• Q₁ = Q₂ = Q₁₂ = 57.6 μC

(c) Voltages:

$V_1 = \frac{Q_1}{C_1} = \frac{57.6}{4.0} = \boxed{14.4 \text{ V}}$

$V_2 = \frac{Q_2}{C_2} = \frac{57.6}{6.0} = \boxed{9.6 \text{ V}}$

$V_3 = \boxed{24 \text{ V}}$

Note: V₁ + V₂ = 14.4 + 9.6 = 24 V ✓

Since capacitor is isolated, charge remains constant: $Q_{before} = Q_{after} = \boxed{5.0 \text{ nC}}$

(b) New voltage:

With dielectric: $C = \kappa C_0 = (2.5)(50) = 125 \text{ pF}$

Voltage decreases: $V = \frac{Q}{C} = \frac{5.0 \times 10^{-9}}{125 \times 10^{-12}}$

$V = \boxed{40 \text{ V}}$

(c) Change in energy:

Initial energy: $U_0 = \frac{1}{2}C_0 V_0^2 = \frac{1}{2}(50 \times 10^{-12})(100)^2 = 2.5 \times 10^{-7} \text{ J}$

Final energy: $U = \frac{1}{2}CV^2 = \frac{1}{2}(125 \times 10^{-12})(40)^2 = 1.0 \times 10^{-7} \text{ J}$

Change in energy: $\Delta U = U - U_0 = 1.0 \times 10^{-7} - 2.5 \times 10^{-7}$

$\Delta U = \boxed{-1.5 \times 10^{-7} \text{ J}}$

Energy decreases (absorbed by dielectric being pulled in).