capacitance capacitors and electric-field energy

From the lesson on potential we know that a potential difference $U = \varphi_1 - \varphi_2$ can be maintained between two conductors, and moving charge from one to the other requires work. Capacitance is the ratio of stored charge to voltage: $C = \frac{Q}{U},$ measured in farads ($1\ \mathrm{F} = 1\ \mathrm{C}/\mathrm{V}$). Capacitance is fixed by geometry and the medium, not by the charge: for a parallel-plate capacitor $C = \frac{\varepsilon_0 \varepsilon S}{d},$ where $S$ is the plate area, $d$ the plate separation and $\varepsilon$ the relative permittivity. Between the plates the field is essentially uniform, $E = U/d$, and the energy of a charged capacitor is $W = \frac{CU^2}{2} = \frac{Q^2}{2C} = \frac{QU}{2}.$ It is convenient to assign this energy to the field itself: in vacuum the volume energy density is $w = \frac{\varepsilon_0 E^2}{2}.$ Finally, capacitors in parallel add their capacitances, while in series the reciprocals add: $C_{\text{eq}} = C_1 + C_2$ and $1/C_{\text{eq}} = 1/C_1 + 1/C_2$.

A parallel-plate capacitor has plate area $S = 1.0\cdot10^{-2}\ \mathrm{m^2}$ and plate separation $d = 2.0\ \mathrm{mm}$; the space between the plates is vacuum ($\varepsilon = 1$). Find: (a) the capacitance $C$; (b) the charge $Q$ at voltage $U = 200\ \mathrm{V}$; (c) the field magnitude $E$ between the plates.

Two capacitors $C_1 = 6.0\ \mu\mathrm{F}$ and $C_2 = 3.0\ \mu\mathrm{F}$ are connected in series to a source of voltage $U = 90\ \mathrm{V}$. Find the equivalent capacitance, the charge on each capacitor, the voltage across each, and the total stored energy.

A capacitor $C_1 = 10\ \mu\mathrm{F}$ charged to voltage $U_0 = 100\ \mathrm{V}$ is connected in parallel to an uncharged capacitor $C_2 = 10\ \mu\mathrm{F}$. Find the final voltage, the initial and final energies of the system, and explain why the energies do not match.

Safe home practicum on a simulation (no real high voltages).

1. Open the PhET simulation «Capacitor Lab: Basics» and select the mode with adjustable geometry.
2. Fix the source voltage at $U = 1.5\ \mathrm{V}$ and, varying the separation $d$ (for example from $1\ \mathrm{mm}$ to $5\ \mathrm{mm}$), record the measured capacitance $C$ and charge $Q$ in a table.
3. Plot $C$ against $1/d$. Check that the points lie on a straight line and find its slope; compare it with $\varepsilon_0 S$ for the chosen area $S$.
4. At fixed $S$ and $d$, vary $U$, record $Q$ and $W$, plot $W$ against $U^2$ and verify that it is linear with slope $C/2$.
5. Write a conclusion: do the measurements confirm $C = \varepsilon_0 S/d$, $Q = CU$ and $W = CU^2/2$?

Important: do all measurements in the simulation only. Do not build real circuits connected to mains, and if you use real components, stay within safe low-voltage sources (up to $12\ \mathrm{V}$).

Show how capacitor geometry and voltage determine charge, field and stored energy, and link these quantities to the formulas of the lesson.

Diagram + graphs. On the left: a parallel-plate capacitor — two parallel plates of area $S$ separated by $d$, with charges $+Q$ and $-Q$ marked, voltage $U$, and the uniform field $E = U/d$ drawn as equally spaced field lines from the positive plate to the negative one. On the upper right: a graph of $C$ against $1/d$ — a straight line through the origin with slope $\varepsilon_0 S$. On the lower right: a graph of $W$ against $U^2$ — a straight line with slope $C/2$. The graph labels show that plate area and separation set the capacitance, while voltage sets the stored energy at fixed capacitance.

The flash capacitor in a phone has capacitance $C = 100\ \mu\text{F}$ and is charged by a converter to $U = 300\ \text{V}$. Before the shot it stores energy $E = \tfrac{1}{2}CU^2 = \tfrac{1}{2}\cdot 100\cdot 10^{-6}\ \text{F}\cdot (300\ \text{V})^2 = 4.5\ \text{J},$ and the plates carry charge $Q = CU = 30\ \text{mC}$. The pulse lasts $t = 1\ \text{ms}$, so the average discharge power is $P = E/t = 4.5\ \text{kW}$; drawing that from a $3.7\ \text{V}$ battery would need about $1200\ \text{A}$, which is impossible.

The scenario uses the energy formula for a charged capacitor $E = \tfrac{1}{2}CU^2 = \tfrac{1}{2}QU$ together with $Q = CU$: the energy is stored in the electric field between the plates, which is why it can be released in milliseconds as a kilowatt power pulse.

Under a smartphone's glass lies a grid of transparent electrodes; the mutual capacitance of one electrode pair is $C_0 \approx 10\ \text{pF}$ and they are held at $U = 1.8\ \text{V}$. The charge of that pair is $Q = C_0U = 10\cdot 10^{-12}\ \text{F}\cdot 1.8\ \text{V} = 18\ \text{pC}$. A finger brought close changes the capacitance by about $\Delta C \approx 1\ \text{pF}$, that is by $10\%$, and the controller registers a charge change $\Delta Q \approx 1.8\ \text{pC}$ at the same $U$.

The relation $Q = CU$ and the dependence of capacitance on geometry and dielectric, $C = \varepsilon\varepsilon_0 S/d$, both apply: a finger is a conductor and its moist tissue is a dielectric with $\varepsilon \approx 40$, so the field between the electrodes is distorted, the pair's capacitance drops, and the sensor measures the recharging $\Delta Q$ at a known voltage $U$.

A cloud and the ground can be approximated by a parallel-plate capacitor: the cloud base has area $S = 1\ \text{km}^2 = 10^6\ \text{m}^2$ and the separation is $d = 1\ \text{km} = 10^3\ \text{m}$, so $C = \frac{\varepsilon_0 S}{d} = \frac{8.85\cdot 10^{-12}\cdot 10^6}{10^3} \approx 8.9\ \text{nF}.$ Before the discharge the field reaches $E = 10^5\ \text{V/m}$, hence the voltage is $U = Ed = 10^8\ \text{V}$, the charge is $Q = CU \approx 0.9\ \text{C}$, and the stored energy is $\tfrac{1}{2}CU^2 \approx 44\ \text{MJ}$ — as much as an electric car battery holds.

Both the definition of capacitance $C = Q/U$ and the electric-field energy $\tfrac{1}{2}QU$ are applied; the field between cloud and ground is nearly uniform, and those surfaces are nearly equipotential, which is what allows the system to be treated as a parallel-plate capacitor.