An interactive companion to the handwritten lecture notes. A capacitor is two isolated conductors that hold +q and −q; capacitance tells you how much charge it stores per volt. Start by charging one below.
Created by Claude Opus 5.5 based on the lecture notes by Prof. Sang Hoon Lee, Department of Physics, Gyeongsang National University.
A capacitor stores electric energy in the form of separated charge, then releases it on demand, which is exactly what a camera flash does. The amount of charge is proportional to the potential difference across it:
The constant C is the capacitance. Its unit is the coulomb per volt, called the farad (F). Crucially, C depends only on the geometry of the plates, not on q or V, so it is a true proportionality constant.
Close the switch and the battery pushes charge onto the plates until the capacitor's voltage matches the battery's. Open it and the charge stays put. Then fire the lamp.
Charging here is drawn as a smooth curve; the real time dependence comes later, once resistance enters the circuit (RC circuits).
Yellow dots show the direction of conventional current. No charge crosses the gap between the plates.
Every capacitance calculation in these notes follows the same loop: assume a charge, find the field it makes, integrate the field to get a potential difference, then divide.
Pick a geometry, drag the dimensions, and watch the capacitance respond. The derivation underneath follows the four steps above.
If A is large, each plate behaves like an infinite conducting sheet, so E is uniform. A Gauss surface around the top plate gives ε0EA = q, so E = q/ε0A. Integrating across the gap, V = Ed. Then ε0EA = C(Ed).
A cylindrical Gauss surface of radius r and length L gives ε0E(2πrL) = q, so E = q/2πε0Lr. Integrating from b to a: V = (q/2πε0L) ln(b/a), using −ln(a/b) = ln(b/a). A coaxial cable is this shape.
A spherical Gauss surface gives ε0E(4πr²) = q, which is just Coulomb's law. Then V = (q/4πε0)(1/a − 1/b). Letting b → ∞ means the outer shell is infinitely far away, and ab/(b−a) → a.
Any network of capacitors can be replaced by a single equivalent capacitance Ceq that draws the same charge from the battery.
Parallel: every capacitor sees the same V, so the charges add. Series: the plates between neighbouring capacitors are isolated, so charge is conserved there: if one side is +q the other must be −q. Every capacitor carries the same q, and the voltages add, like elevation gains along a path up a hill.
Watch out: this is the reverse of resistors, which you will meet later. Resistors add in series and add as reciprocals in parallel.
Charging a capacitor takes work. With charge q′ already on it, moving one more sliver dq′ costs dW = V′dq′ = (q′/C)dq′. The more charge has piled up, the higher V′ climbs, so each slice gets harder, in exact proportion to q′. Adding up the slices is the area under the line:
More slices make the staircase converge on the triangle, which is the integral.
In a parallel-plate capacitor the field is uniform, so the energy per unit volume is uniform too. Dividing by the volume Ad:
u = UAd = CV²2Ad = 12ε0(V/d)²
This last form holds in general, not just between plates: wherever there is an electric field, there is potential energy stored in space at this density.
Insulating materials such as oil or plastic are called dielectrics. Filling the gap with one multiplies the capacitance by the dielectric constant κ (kappa):
Here ℒ is a length set by geometry (A/d for parallel plates). Strictly, κ is measured against vacuum, but air's κ = 1.00054 is close enough. Since q = CV still holds, what happens next depends on whether the battery stays connected.
Fixed setup: A = 100 cm², d = 1 mm, charged to 12 V before any change.
Molecules in the dielectric are electric dipoles (water is a polar molecule). The external field E₀ exerts a torque that lines them up. The aligned dipoles leave a sheet of −q′ just under the + plate and +q′ just above the − plate, and that induced charge makes its own field E′ pointing the other way. The net field is the vector sum, E = E₀ − E′. In a conductor the cancellation is perfect, E′ = −E₀, which is why the field inside a conductor is zero.
Gauss's law inside a dielectric therefore counts only free charge if you multiply by κ:
Comparing with ε0∮E·dA = q − q′ gives q − q′ = q/κ. A point charge in a dielectric produces E = q/4πκε0r², and the field just outside a conductor is σ/κε0: in both cases the vacuum value divided by κ. In the notes this is written with D = κE; the standard SI convention puts the ε₀ inside, D = κε₀E, so that ∮D·dA = qfree.
With q fixed, U falls to U/κ and the lost energy is the work the field does pulling the slab in, exactly as the notes say. With the battery connected, U rises to κU, but the slab is still pulled in, not pushed out. The battery supplies (κ−1)CV² of energy: half of it raises the stored energy and the other half is the work done pulling the slab into the gap. You don't need to force it in either case.
Place a slab of thickness b and constant κ in the middle of a gap d. Between the plate and slab (point i) the field is q/ε0A; inside the slab (point ii) it drops to q/κε0A. The potential falls steeply through air and gently through the slab:
V = qε0A(d − b) + qκε0Ab ⇒
Check the limits: b → 0 gives ε₀A/d, and b = d gives κε₀A/d.
Fixed setup: A = 100 cm², d = 1 mm. Graphs are in units of q/ε₀A (field) and qd/ε₀A (potential); x runs from the + plate to the − plate.