Module 6
Fields and Capacitors
Start with the rubber sheet: it's the picture behind every capacitor flashcard. Then six benches. The first three build the picture of an electric field as a landscape of hills and pits. The last three show why a capacitor charges the way it does, and why everything about it is exponential. Colours are the same throughout:
The rubber sheet in a pipe that's already full.
6.1.1(b)
how a capacitor charges
The rubber sheet in the pipe
Forget the circuit for a minute. Picture a loop of pipe already completely full of water, with a pump on one side and a stretchy rubber sheet across the pipe on the other. Water can never get through the sheet. Press Charge and watch.
1. The pump doesn't make water, it moves it. The pipe is full on both sides of the sheet. A battery is the same: the wires and plates are metal and already full of free electrons. The battery pushes electrons out of its − terminal and pulls the same number in at its + terminal.
2. Nothing crosses the sheet, yet water moves all the way round. Water arriving on one side makes the sheet bulge, and the bulge shoves along the water already sitting on the other side. Same amount in on one side, same amount pushed along on the other, because water can't be squashed. In the circuit: electrons pile onto plate A (it goes negative), and an equal number are pushed off plate B towards the + terminal (it goes positive). That's why an ammeter on either side of a capacitor reads the same current, even though no charge crosses the gap.
3. The stretched sheet pushes back. The more it bulges, the harder it pushes back against the pump. Only the difference between the pump's push and the sheet's push-back drives the flow, so the flow slows down. When the sheet pushes back as hard as the pump, the flow stops, with the pump still on. In the circuit: VC grows until it equals the e.m.f., and the current falls to zero.
4. Discharging is the sheet springing back. Take the pump out and join the pipe. The stretched sheet shoves the water back the other way round, fast at first (biggest push-back), then slower and slower as it relaxes. In the circuit: the spare electrons on plate A flow round through the resistor to plate B, filling the gaps left there; that is all "neutralise" means. Their energy ends up as heat in the resistor.
5. What the sheet really is: the electric field. Electrons can't cross the gap, but their field can. The same extra electrons on plate A do three jobs through it:
- They repel the incoming electrons on A's side. That's the push-back, VC.
- They repel the electrons in plate B across the gap and push them off down the wire. That's why B goes positive by exactly the same amount.
- Once the battery is removed, positive plate B attracts A's extra electrons back across the gap and holds them on A's inner face. The charge stays put because each plate holds the other's charge in place, and there's no route out until you connect a wire.
Between the plates there's a uniform field, \(E = V/d\), pointing from B (+) to A (−), and the stored energy lives in it. Discharging is that field relaxing as the electrons flow round through the resistor. Real capacitors also leak slowly through the insulator, which is why one left on its own eventually goes flat.
It's tempting to picture a hose being clamped shut: water piles up behind the clamp and nothing happens on the far side. That picture is wrong for a capacitor, because it leaves the far side empty and still. The pipe is full on both sides, and the bulge pushes the far side's water along. That's where the positive plate comes from: it isn't "given" positive charge, its electrons are pushed away.
| Water picture | Capacitor circuit |
|---|---|
| water (pipe already full) | free electrons (metal already full) |
| pump | battery (e.m.f.) |
| narrow section | resistor |
| rubber sheet | insulator between the plates; its stretch is the electric field across the gap |
| water pressed against the sheet | plate A: extra electrons, −Q |
| water pushed away on the far side | plate B: electrons removed, +Q |
| how hard the sheet pushes back | p.d. across the capacitor, VC |
| stretch energy in the sheet | energy stored, ½QV |
Electric fields
6.2.1, 6.2.4(a)
Potential is a landscape; the field is the slope
Think of electric potential as height. A positive charge is a hill (red shading), a negative charge is a pit (blue). A positive test charge rolls downhill: that's the force, and the field lines show which way. Where the shading changes fast, the slope is steep and the field is strong. Drag the charges, and move your pointer to place a test charge.
- Electric field strength: the force per unit positive charge, \(E = F/Q\). A vector.
- Electric potential: the work done per unit positive charge in bringing a positive charge from infinity to the point. A scalar: potentials just add.
- Field lines show the direction of the force on a positive charge, meet conductors at 90°, and never cross.
6.2.2, 6.2.4(b)(d)
1/r² for the field, 1/r for the potential
Move away from a point charge. The field falls as \(1/r^2\) (the slope flattens fast); the potential falls as \(1/r\) (the height drops more slowly). The shaded area under the force-distance curve is the work done bringing the charge in from the far end, and it equals q × the potential difference. Double r and watch: E drops to a quarter, V to a half.
- \(E = Q/(4\pi\varepsilon_0 r^2)\) and \(V = Q/(4\pi\varepsilon_0 r)\). Double r: E is ¼, V is ½.
- Area under a force-distance graph = work done.
- \(E = V/d\) is only for a uniform field. Using it with a point charge "is wrong physics and scores zero".
6.2.3
Between parallel plates the slope is constant
Between two plates the potential drops steadily from one plate to the other, so the slope (the field) is the same everywhere: \(E = V/d\). A charge fired in sideways is like a ball thrown horizontally: the sideways speed never changes, while a constant force pulls it across, so the path is a parabola. Switch to a proton: same size force, but about 1840 times the mass, so it barely bends.
- Horizontal: no force, constant speed, \(t = L/v\). Vertical: \(a = EQ/m\), \(v_y = at\). Path is a parabola.
- Energy gained crossing the plates = QV. Changing d doesn't change it: F halves when d doubles, but the distance doubles.
- Parallel-plate capacitor: \(C = \varepsilon_0 A/d\). Field lines between the plates: parallel, equally spaced, at 90° to the plates.
Capacitors
6.1.1(b), 6.1.3(a)(b)
Why charging slows down
At the start the capacitor is empty, so it has no p.d.: all 6 V sits across the resistor and a big current flows. Electrons pile onto one plate and leave the other in equal numbers. As charge builds up, the capacitor's own p.d. pushes back, leaving less p.d. across the resistor, so the current gets smaller, so it charges more slowly. That feedback (the rate depends on how much is left to go) is exactly what makes a curve exponential. Watch the electrons slow down.
- Charging: electrons are deposited on one plate and an equal number removed from the other; none cross the insulator.
- At every instant \(V_0 = V_C + V_R\). As VC rises, \(V_R = IR\) falls, so I falls.
- Time constant: the time for the p.d. / current / charge to fall to 1/e (37%) of its initial value. Not "CR" on its own. \(\tau = CR\).
- Discharging: energy is dissipated as heat in the resistor.
6.1.3(d)(e)
The same fraction lost every step
Discharging: the current is \(Q/CR\), so in a short time Δt the capacitor loses a fixed fraction \(\Delta t/CR\) of whatever charge it has left. Lose the same fraction again and again and you get an exponential, with the constant-ratio property: equal time steps, equal ratios. Press Step to run the spreadsheet model by hand. Make the step smaller and the staircase hugs the true curve.
| step | t / CR | Q left (%) | lost this step | ratio Qₙ₊₁/Qₙ |
|---|
- Iterative model: \(\Delta Q = (\Delta t/CR) \times Q\), then \(Q_{\text{new}} = Q - \Delta Q\). Choose Δt much smaller than CR.
- Constant ratio: "for fixed intervals of t, V falls by the same ratio". V halves every 0.69CR; falls to 0.37 every CR.
- Straight line: \(\ln V = \ln V_0 - t/CR\), so plot ln V against t: gradient = \(-1/CR\).
6.1.2
Why the energy is ½QV, not QV
Each new bit of charge has to be pushed onto a plate that is already charged, against the p.d. it has built up so far. The first bit costs almost nothing; the last bit is pushed against nearly the full V. Add charge in chunks: each strip is the work for one chunk (p.d. × charge). Together they fill the triangle under the V-Q graph, which is \(\tfrac12 QV\). Use more chunks and the strips fit the triangle exactly.
Extra insight: the battery delivers charge Q at its full e.m.f., so it supplies QV. The capacitor keeps \(\tfrac12 QV\). The other half is turned into heat in the resistor while charging, whatever the resistance.
- Area under the p.d.-charge graph = energy stored. Gradient of V against Q = \(1/C\).
- \(W = \tfrac12 QV = \tfrac12 CV^2 = \tfrac12 Q^2/C\). Double the p.d.: energy ×4.
- Watch PMT's definitions list: it says "charge × capacitance". It's charge × p.d..