Unit 4

Capacitance & Dielectrics

A capacitor is two conductors separated by an insulator: the simplest device that stores charge and energy in an electric field. It is also one of the three passive components every circuit is built from (with resistors and inductors), and there are probably more than a thousand of them on the motherboard of the device you are reading this on. This unit puts Units 2 and 3 to work: Gauss's law gives the field between the conductors, the potential integral gives the voltage, and their ratio is the capacitance. Then we fill the gap with an insulator — a dielectric — and see why that multiplies the capacitance. The lab lets you slide a dielectric slab into a capacitor, with the battery connected or not, and watch \(Q\), \(V\), \(E\) and the stored energy respond.

1 · What a capacitor is

Take two separate conductors (often called plates) and move charge \(Q\) from one to the other, so one holds \(+Q\) and the other \(-Q\). A battery does exactly this when you connect it. An electric field now runs from the \(+Q\) conductor to the \(-Q\) one, and there is a potential difference \(V\) between them.

Here is the key fact: double \(Q\) and every field vector in space doubles (superposition), so the potential difference, an integral of the field, doubles too. \(Q\) and \(V\) are proportional. The constant of proportionality is the capacitance:

\[ C = \frac{Q}{V}\qquad\qquad 1\ \text{farad (F)} = 1\ \text{C/V} \] \(Q\) is the charge on one conductor (the magnitude, not the net charge, which is zero); \(V\) is the magnitude of the potential difference between them. \(C\) measures how much charge the device holds per volt. A farad is enormous: real capacitors are typically pF (\(10^{-12}\) F) to mF (\(10^{-3}\) F).

Common mistake

Thinking \(C\) depends on \(Q\) or \(V\) because it is written \(Q/V\). It does not: \(C\) depends only on geometry and the insulating material. Raising the voltage raises the charge in proportion; the ratio stays fixed — just as a resistor's \(R = V/I\) does not change when you raise \(V\).

EE connection

On a schematic a capacitor is two parallel lines, a literal picture of the device. Engineers use them to decouple power rails (a 100 nF capacitor next to every chip supplies sudden current spikes), to filter (with resistors they make the RC circuits of Unit 5), to store energy (camera flashes, defibrillators), and to store information: each bit in a DRAM chip is a capacitor of a few tens of femtofarads, charged or not.

2 · The recipe for computing \(C\)

Every capacitance derivation in this course follows the same four steps:

Assume charges \(+Q\) and \(-Q\) on the two conductors.
Find \(\mathbf{E}\) in the gap, almost always with Gauss's law (Unit 2), using the symmetry of the shape.
Find \(V\) by integrating from one conductor to the other: \(V = \left|\int_-^+\mathbf{E}\cdot d\boldsymbol{\ell}\right|\) (Unit 3).
Divide: \(C = Q/V\). \(Q\) must cancel. If it does not, there is an algebra error.

The units check

Every vacuum capacitance comes out as \(\varepsilon_0\times(\text{a length})\), and \(\varepsilon_0 = 8.85\ \text{pF/m}\). So a capacitor about a centimetre in size, with air between its conductors, is roughly a picofarad or less. Big capacitance needs huge area, tiny gaps, or a dielectric — usually all three.

3 · Parallel plates, coaxial cylinders, spheres

Parallel-plate capacitor

Between large, close plates the field is uniform, \(E = \sigma/\varepsilon_0\), and zero outside (the two sheets' fields cancel there). The bulging “fringe” field at the edges is ignored in the ideal formula.

Plates of area \(A\), separation \(d\), with \(d\) much smaller than the plate width. Gauss's law (Unit 2): \(E = \sigma/\varepsilon_0 = Q/(\varepsilon_0A)\), uniform. The field is uniform, so \(V = Ed\):

\[ V = Ed = \frac{Qd}{\varepsilon_0 A}\qquad\Longrightarrow\qquad C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d} \] Bigger area, more capacitance (more room for charge at a given voltage). Smaller gap, more capacitance (a given charge produces less voltage across a shorter distance). Example: 1 cm² plates 1 mm apart give \(C = (8.85\times10^{-12})(10^{-4})/(10^{-3}) = 0.885\) pF.

Cylindrical (coaxial) capacitor

A coaxial cable seen end-on. Gauss's law on the dashed cylinder (length \(L\)) encloses only the inner conductor's charge, giving the field of a line charge.

Inner conductor radius \(a\), outer radius \(b\), length \(L\), charge \(\pm Q\), so \(\lambda = Q/L\). Between them \(E = \lambda/(2\pi\varepsilon_0 r)\). Integrate from \(a\) to \(b\) (Unit 3's line-charge potential):

\[ V = \int_a^b\frac{\lambda}{2\pi\varepsilon_0 r}\,dr = \frac{\lambda}{2\pi\varepsilon_0}\ln\frac{b}{a}\qquad\Longrightarrow\qquad C = \frac{Q}{V} = \frac{\lambda L}{V} = \frac{2\pi\varepsilon_0 L}{\ln(b/a)} \] Proportional to length, so cables are rated in capacitance per metre, \(C/L = 2\pi\varepsilon_0/\ln(b/a)\). Only the ratio \(b/a\) matters: a thick cable and a thin one with the same proportions have the same pF/m.

EE connection

The coax that carries a radio or oscilloscope signal is this capacitor. Typical 50 Ω coax has roughly 100 pF per metre (Worked example 2), so a 2 m cable adds about 200 pF of load to whatever drives it. That is why a scope probe has a compensation adjustment, and why fast digital signals need properly terminated transmission lines (Unit 9's waves) rather than “just a wire”.

Spherical capacitor, and the isolated sphere

A sphere of radius \(a\) inside a concentric shell of radius \(b\). Between them the field is that of a point charge, \(E = kQ/r^2\):

\[ V = \int_a^b\frac{kQ}{r^2}\,dr = kQ\left(\frac1a - \frac1b\right) = kQ\,\frac{b-a}{ab}\qquad\Longrightarrow\qquad C = \frac{ab}{k(b-a)} = 4\pi\varepsilon_0\frac{ab}{b-a} \] Two checks. Let \(b\to\infty\): \(C\to4\pi\varepsilon_0 a\), the capacitance of an isolated sphere (its partner conductor is “at infinity”). Let \(b - a = d\) be tiny: \(C\approx4\pi\varepsilon_0a^2/d = \varepsilon_0A/d\), the parallel-plate formula, because a thin shell gap looks flat up close.

The whole Earth, as an isolated sphere (\(R = 6.37\times10^6\) m), has \(C = R/k = 6.37\times10^{6}/8.99\times10^{9} \approx 7.1\times10^{-4}\ \text{F} = 710\ \mu\text{F}\) — less than a capacitor you can buy for well under a dollar. The farad really is huge.

Geometry	Field in the gap	Capacitance (vacuum)
Parallel plates (area \(A\), gap \(d\))	\(\sigma/\varepsilon_0\), uniform	\(\varepsilon_0A/d\)
Coaxial cylinders (radii \(a<b\), length \(L\))	\(\lambda/(2\pi\varepsilon_0r)\)	\(2\pi\varepsilon_0L/\ln(b/a)\)
Concentric spheres (radii \(a<b\))	\(kQ/r^2\)	\(4\pi\varepsilon_0\,ab/(b-a)\)
Isolated sphere (radius \(R\))	\(kQ/r^2\) outside	\(4\pi\varepsilon_0R\)

4 · Capacitors in series and parallel

Parallel capacitors share a voltage; series capacitors share a charge. The middle wire in the series chain is isolated, so whatever charge leaves one plate appears on the other: \(-Q\) and \(+Q\) on the two plates it connects.

\[ \text{Parallel:}\quad C_{\text{eq}} = C_1 + C_2 + \cdots\qquad\qquad \text{Series:}\quad \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \] Parallel: same \(V\) across each, total charge \(Q = C_1V + C_2V\), so the capacitances add — like making the plate area bigger. Series: same \(Q\) on each, total voltage \(V = Q/C_1 + Q/C_2\), so the reciprocals add — like making the gap bigger. A series combination is always smaller than its smallest member.

Common mistake

The rules are the opposite of resistors (resistors add in series, capacitors add in parallel). Two equal capacitors in series give \(C/2\), in parallel \(2C\). Another frequent slip: forgetting to take the reciprocal at the end of a series calculation — \(1/C_{\text{eq}} = 1/4 + 1/12 = 1/3\) means \(C_{\text{eq}} = 3\), not \(1/3\).

Strategy for networks

Reduce the network step by step to one equivalent capacitor, find the total charge \(Q = C_{\text{eq}}V\), then expand back out: in series pieces use “same \(Q\)” to find each \(V\); in parallel pieces use “same \(V\)” to find each \(Q\). Check that voltages around each loop add up.

5 · Stored energy and energy density

Charging a capacitor takes work: each new bit of charge \(dq\) must be pushed onto a plate that already repels it. When the capacitor holds \(q\), its voltage is \(v = q/C\), and moving \(dq\) across costs \(dW = v\,dq\). Add up from 0 to \(Q\):

\[ U = \int_0^Q\frac{q}{C}\,dq = \frac{Q^2}{2C} = \frac12 CV^2 = \frac12 QV \] All three forms are equal (substitute \(Q = CV\)); use whichever has the quantities you know. The \(\tfrac12\) is because the voltage grows from 0 to \(V\) as you charge: the average voltage during charging was \(V/2\).

Voltage rises linearly with charge (slope \(1/C\)). The work to charge is the area under the line, a triangle: \(\tfrac12QV\). Compare a battery, which delivers charge at a nearly constant voltage.

Where is the energy? In the field.

For a parallel-plate capacitor, \(C = \varepsilon_0A/d\) and \(V = Ed\): \(U = \tfrac12(\varepsilon_0A/d)(Ed)^2 = \tfrac12\varepsilon_0E^2\,(Ad)\). And \(Ad\) is the volume of the gap, where the field lives. Divide by it:

\[ u = \frac{U}{\text{volume}} = \frac12\varepsilon_0E^2\qquad(\text{J/m}^3) \] The energy density of the electric field. Derived here for one shape, but true everywhere: wherever there is an electric field, there is energy \(\tfrac12\varepsilon_0E^2\) per cubic metre. In Unit 9 this is the energy an electromagnetic wave carries through empty space.

A sense of scale: at air's breakdown field, \(3\times10^6\) V/m, \(u = \tfrac12(8.85\times10^{-12})(3\times10^6)^2\approx40\ \text{J/m}^3\) — tiny compared with a battery's chemical energy (around a billion joules per cubic metre for a Li-ion cell). That is why capacitors deliver energy fast but do not store much.

EE connection

A camera flash charges a capacitor of about 100 μF to 330 V: \(U = \tfrac12(100\times10^{-6})(330)^2\approx5.4\) J, released in about a millisecond — thousands of watts that the small battery could never supply directly. A defibrillator works the same way at higher energy. Supercapacitors push capacitance to thousands of farads using electrode surfaces with enormous area and molecular-scale gaps.

6 · Dielectrics

Fill the gap with an insulating material (a dielectric) and the capacitance goes up by a factor \(\kappa\) (kappa), the material's dielectric constant:

\[ C = \kappa C_0,\qquad\text{e.g.}\quad C = \frac{\kappa\varepsilon_0A}{d} = \frac{\varepsilon A}{d},\qquad \varepsilon = \kappa\varepsilon_0 \] \(C_0\) is the capacitance with vacuum in the gap. \(\kappa\ge1\) is dimensionless; \(\varepsilon\) is the material's permittivity. Every vacuum formula in Section 3 becomes the dielectric-filled version by replacing \(\varepsilon_0\) with \(\kappa\varepsilon_0\) (when the dielectric fills the whole field region).

A dielectric does three useful things: it raises \(C\); it holds the plates apart mechanically; and most dielectrics withstand a much higher field than air before breaking down (becoming conducting and arcing through). The maximum field a material survives is its dielectric strength \(E_{\max}\). It sets the capacitor's voltage rating: \(V_{\max} = E_{\max}d\).

Typical room-temperature values from standard textbook tables; real materials vary with purity, frequency and temperature.
Material	\(\kappa\)	Dielectric strength (MV/m)
Vacuum	1 (exactly)	—
Air (dry, 1 atm)	1.00059	3
Teflon	2.1	60
Polyethylene	2.3	18
Polystyrene	2.56	24
Mylar (polyester film)	3.2	7
Paper	3.7	16
Silicon dioxide (chip insulator)	3.9	≈ 1000 (thin films)
Pyrex glass	5.6	14
Water (20 °C)	80	—
Strontium titanate	233	8

Interactive lab Dielectrics & energy density Open it in Interactive Laboratories →

EE connection

Capacitor types are named after their dielectric: ceramic (MLCC), film, electrolytic (an oxide layer only nanometres thick), tantalum. Class-2 ceramics use barium-titanate materials with \(\kappa\) in the thousands, which is how a 10 μF part fits in a 1 mm package — but their \(\kappa\) drops as the applied voltage rises, so a “10 μF” MLCC can deliver much less at its rated voltage. Datasheets call this DC-bias derating, and it bites every new board designer once. Inside a chip, the MOSFET gate is a capacitor too: SiO₂ (\(\kappa = 3.9\)) a couple of nanometres thick gives \(C/A = \kappa\varepsilon_0/t\approx0.017\ \text{F/m}^2\), about 17 fF per square micrometre.

7 · What happens inside a dielectric

Why does an insulator increase capacitance? An insulator has no free charges, but its molecules can still be distorted or rotated. In a field, each molecule's positive part is pushed slightly along \(\mathbf{E}\) and its negative part slightly against it: the molecule becomes a tiny dipole aligned with the field. This is polarization. Some molecules (like water) are permanent dipoles that the field rotates into line; others are stretched into dipoles. Either way the result is the same.

Aligned molecular dipoles cancel each other inside the slab, but leave uncancelled bound charge on its two faces: negative next to the positive plate, positive next to the negative plate. Their field opposes the plates' field, so the field inside the dielectric is reduced to \(E_0/\kappa\).

Deep inside, each dipole's \(+\) end sits next to its neighbour's \(-\) end, so the interior stays neutral. But at the faces the cancellation fails: a thin layer of bound charge appears (bound, because it cannot leave its molecules). Its field points against the applied field and partly cancels it. For a dielectric filling the gap:

\[ E = \frac{E_0}{\kappa},\qquad \sigma_{\text{b}} = \sigma\left(1 - \frac{1}{\kappa}\right) \] \(E_0 = \sigma/\varepsilon_0\) is the field the free charge alone would make; \(\sigma_{\text{b}}\) is the magnitude of the bound surface charge density. The net charge density at each face is \(\sigma - \sigma_{\text{b}} = \sigma/\kappa\), hence \(E = \sigma/(\kappa\varepsilon_0)\). With less field for the same free charge, the voltage is lower, and \(C = Q/V\) is larger by \(\kappa\).

Breakdown is what happens when the field is strong enough to rip electrons free of their molecules: the few free electrons accelerate, knock more loose, and an avalanche turns the insulator into a conducting path. In a solid this usually burns a permanent hole; in air it is a spark that heals itself.

Energy density in a dielectric

With a dielectric, \(u = \tfrac12\kappa\varepsilon_0E^2\). Because good dielectrics survive far higher fields than air, they can store far more energy per volume: polystyrene at 24 MV/m holds about 6.5 kJ/m³, roughly 160 times air's maximum.

8 · Connected or isolated: what changes?

A classic exam question: change something about a charged capacitor (insert a dielectric, pull the plates apart) and predict \(Q\), \(V\), \(E\) and \(U\). The one thing you must decide first is which quantity is held fixed:

Still connected to the battery: the battery holds \(V\) fixed. Charge flows on or off the plates as needed.
Charged, then disconnected (isolated): the charge has nowhere to go, so \(Q\) is fixed. The voltage changes.

Change	Condition	\(C\)	\(Q\)	\(V\)	\(E\)	\(U\)
Insert dielectric \(\kappa\) (fills gap)	Isolated (\(Q\) fixed)	\(\times\kappa\)	same	\(\div\kappa\)	\(\div\kappa\)	\(\div\kappa\)
Insert dielectric \(\kappa\) (fills gap)	Connected (\(V\) fixed)	\(\times\kappa\)	\(\times\kappa\)	same	same	\(\times\kappa\)
Double the gap \(d\)	Isolated (\(Q\) fixed)	\(\div2\)	same	\(\times2\)	same	\(\times2\)
Double the gap \(d\)	Connected (\(V\) fixed)	\(\div2\)	\(\div2\)	same	\(\div2\)	\(\div2\)

Each row comes from two formulas: \(C\) from the geometry, then \(U = Q^2/2C\) if \(Q\) is fixed or \(U = \tfrac12CV^2\) if \(V\) is fixed — always pick the form whose variable is held constant. For the field, \(E = V/d\) in a parallel-plate capacitor (it covers both cases).

Where the energy goes

Isolated capacitor, dielectric inserted: \(U\) drops. The lost energy did work on the slab — the fringe field pulls a dielectric into a capacitor. Isolated capacitor, plates pulled apart: \(U\) rises, supplied by your hands working against the plates' attraction. Connected capacitor: the battery also does work, \(W_{\text{bat}} = V\,\Delta Q\), which is twice the change in stored energy when a dielectric is inserted; the other half does work on the slab. Do not expect “energy stored” to be conserved by itself when a battery is attached.

EE connection

Capacitive sensing is this table in action. A touchscreen senses your finger changing the capacitance of electrodes under the glass. A MEMS accelerometer in a phone measures a tiny moving plate's change in gap \(d\). A capacitive level sensor measures how far a liquid with large \(\kappa\) fills the space between two electrodes. In every case the electronics measure \(C\), and the physics links it to position or material.

9 · Worked examples

1. A parallel-plate capacitor: C, Q, E and U

Plates 2.0 cm by 3.0 cm, 1.0 mm apart in air, connected to 12 V. Find \(C\), \(Q\), \(E\) and \(U\), and check \(U\) using the energy density.

\[ A = (0.020)(0.030) = 6.0\times10^{-4}\ \text{m}^2,\qquad C = \frac{\varepsilon_0A}{d} = \frac{(8.85\times10^{-12})(6.0\times10^{-4})}{1.0\times10^{-3}} = \mathbf{5.31\ pF} \] \[ Q = CV = (5.31\times10^{-12})(12) = \mathbf{63.7\ pC},\qquad E = \frac{V}{d} = \frac{12}{1.0\times10^{-3}} = \mathbf{1.2\times10^{4}\ V/m} \] \[ U = \tfrac12CV^2 = \tfrac12(5.31\times10^{-12})(144) = \mathbf{3.82\times10^{-10}\ J} \] \[ \text{Check: } u\cdot(\text{volume}) = \tfrac12(8.85\times10^{-12})(1.2\times10^{4})^2\times(6.0\times10^{-7}\ \text{m}^3) = 3.82\times10^{-10}\ \text{J}\ \checkmark \]

2. Capacitance per metre of a coaxial cable

A coax has inner conductor radius 0.45 mm, outer conductor (inner surface) radius 1.47 mm, and polyethylene insulation (\(\kappa = 2.3\)). Find its capacitance per metre and the capacitance of a 2.0 m length.

\[ \ln\frac{b}{a} = \ln\frac{1.47}{0.45} = \ln 3.267 = 1.184 \] \[ \frac{C}{L} = \frac{2\pi\kappa\varepsilon_0}{\ln(b/a)} = \frac{2\pi(2.3)(8.85\times10^{-12})}{1.184} = \mathbf{1.08\times10^{-10}\ F/m = 108\ pF/m} \] \[ C_{2\,\text{m}} = (108\ \text{pF/m})(2.0\ \text{m}) = \mathbf{216\ pF} \]

In the same range as the roughly 100 pF/m printed on common 50 Ω coax datasheets. Without the polyethylene (air) it would be \(108/2.3 = 47\) pF/m.

3. Reducing a capacitor network

\(C_1 = 4.0\ \mu\text{F}\) and \(C_2 = 12\ \mu\text{F}\) are in series; that pair is in parallel with \(C_3 = 2.0\ \mu\text{F}\); the whole network is across 24 V. Find \(C_{\text{eq}}\), the charge and voltage of each capacitor, and the total stored energy.

\[ C_{12} = \left(\frac{1}{4.0} + \frac{1}{12}\right)^{-1} = \left(\frac{3 + 1}{12}\right)^{-1} = 3.0\ \mu\text{F},\qquad C_{\text{eq}} = 3.0 + 2.0 = \mathbf{5.0\ \mu F} \]

The series pair and \(C_3\) both sit across the full 24 V (they are in parallel):

\[ Q_1 = Q_2 = C_{12}(24) = \mathbf{72\ \mu C},\qquad V_1 = \frac{72}{4.0} = \mathbf{18\ V},\qquad V_2 = \frac{72}{12} = \mathbf{6.0\ V} \] \[ Q_3 = (2.0)(24) = \mathbf{48\ \mu C},\qquad V_3 = \mathbf{24\ V} \] \[ U = \tfrac12C_{\text{eq}}V^2 = \tfrac12(5.0\times10^{-6})(24)^2 = \mathbf{1.44\ mJ} \]

Check: \(V_1 + V_2 = 24\) V ✓. Note the smaller series capacitor takes the larger share of the voltage — useful (and dangerous) when stacking capacitors to reach a higher voltage rating.

4. Inserting a dielectric: isolated vs connected

An air capacitor \(C_0 = 20\) pF is charged to 100 V. A slab with \(\kappa = 4.0\) is then slid in, filling the gap. Find the final \(Q\), \(V\) and \(U\) if (a) the battery was disconnected first; (b) the battery stays connected.

Initially \(Q_0 = C_0V_0 = 2.0\) nC and \(U_0 = \tfrac12C_0V_0^2 = \tfrac12(20\times10^{-12})(100)^2 = 1.0\times10^{-7}\) J. Afterwards \(C = \kappa C_0 = 80\) pF.

\[ \text{(a) } Q \text{ fixed:}\quad Q = \mathbf{2.0\ nC},\quad V = \frac{Q}{C} = \frac{2.0\times10^{-9}}{80\times10^{-12}} = \mathbf{25\ V},\quad U = \frac{Q^2}{2C} = \frac{(2.0\times10^{-9})^2}{2(80\times10^{-12})} = \mathbf{2.5\times10^{-8}\ J} \] \[ \text{(b) } V \text{ fixed:}\quad V = \mathbf{100\ V},\quad Q = CV = \mathbf{8.0\ nC},\quad U = \tfrac12CV^2 = \mathbf{4.0\times10^{-7}\ J} \]

Energy bookkeeping in (b): the battery pushed \(\Delta Q = 6.0\) nC through 100 V, doing \(W = 6.0\times10^{-7}\) J. Stored energy rose by \(3.0\times10^{-7}\) J; the other \(3.0\times10^{-7}\) J was work done pulling the slab in. In (a), \(U\) fell by \(7.5\times10^{-8}\) J, also spent pulling the slab in. Both match the table in Section 8 (\(U\div4\) and \(U\times4\)).

5. Designing for a voltage rating

Design a 1.0 nF parallel-plate capacitor rated for 1.0 kV using paper (\(\kappa = 3.7\), strength 16 MV/m), with a safety factor of 2 on the field. Find the gap and plate area.

Safety factor 2 means the working field may be at most \(16/2 = 8.0\) MV/m.

\[ d = \frac{V}{E_{\text{allowed}}} = \frac{1000}{8.0\times10^{6}} = \mathbf{1.25\times10^{-4}\ m = 0.125\ mm} \] \[ A = \frac{Cd}{\kappa\varepsilon_0} = \frac{(1.0\times10^{-9})(1.25\times10^{-4})}{(3.7)(8.85\times10^{-12})} = \frac{1.25\times10^{-13}}{3.27\times10^{-11}} = \mathbf{3.8\times10^{-3}\ m^2 \approx 38\ cm^2} \]

A 6 cm square — awkward. Real film capacitors roll long strips of foil and dielectric into a cylinder to fit that area into a small can. Notice the trade-off: a higher voltage rating needs a thicker gap, which needs more area for the same \(C\).

6. Charge sharing, and the missing energy

\(C_1 = 6.0\ \mu\text{F}\) is charged to 12 V, disconnected, then connected (+ to +) across an uncharged \(C_2 = 3.0\ \mu\text{F}\). Find the final voltage and the energy before and after.

Charge is conserved on the connected plates; afterwards both share one voltage (they are in parallel).

\[ Q = C_1V_0 = (6.0)(12) = 72\ \mu\text{C},\qquad V_f = \frac{Q}{C_1 + C_2} = \frac{72}{9.0} = \mathbf{8.0\ V} \] \[ U_i = \tfrac12(6.0\times10^{-6})(12)^2 = \mathbf{432\ \mu J},\qquad U_f = \tfrac12(9.0\times10^{-6})(8.0)^2 = \mathbf{288\ \mu J} \]

One third of the energy, 144 μJ, is gone: dissipated as heat in the wires (and a little radiated) while the current flowed. Remarkably the loss does not depend on the wire resistance — a smaller resistance just means a bigger, shorter current pulse. Charge is conserved; stored energy is not.

EE connection

This loss is fundamental to switched-capacitor circuits and to CMOS logic: every time a gate charges its load capacitance \(C\) from a supply \(V\), the supply delivers \(CV^2\) but only \(\tfrac12CV^2\) is stored; the rest is heat. That is the origin of the famous dynamic power formula \(P\propto CV^2f\), and why lowering the supply voltage saves so much power.

Checklist

I know \(C = Q/V\) depends only on geometry and material, never on \(Q\) or \(V\).
I can derive \(C\) for plates, coax and spheres: assume \(\pm Q\), Gauss for \(E\), integrate for \(V\), divide.
I combine capacitors correctly (parallel adds, series adds reciprocals) and expand back to each \(Q\) and \(V\).
I use \(U = Q^2/2C = \tfrac12CV^2 = \tfrac12QV\) and \(u = \tfrac12\varepsilon_0E^2\).
I can explain polarization and bound charge, and why \(E\) drops by \(\kappa\) inside a dielectric.
I check dielectric strength: \(V_{\max} = E_{\max}d\).
Before any “what changes” question, I decide whether \(Q\) or \(V\) is held fixed.