Lesson 1 of 11 · 6 min

The charging capacitor puzzle

NCERT §8.1, §8.2

Kavya charges her capacitor with 0.20 A and holds a compass near the lead. The needle swings. Now she asks: what does Ampere's law say about the field next to the gap, where no wire runs?

The story this chapter follows: Kavya's house full of waves

Kavya's physics kit has a parallel-plate capacitor, plates of radius 4.0 cm, that she charges with a steady 0.20 A. Around her, the house is full of waves she cannot see: an old FM radio tuned to 100 MHz, a phone, a microwave oven, a TV remote, sunlight through the window, a UV water purifier in the kitchen, and a chest X-ray waiting at the clinic. Every idea in this chapter is tried out on them.

The lesson in notes

In short

A current produces a magnetic field (Chapter 4), and a magnetic field that changes with time produces an electric field (Chapter 6). Maxwell argued that the converse also holds: an electric field that changes with time produces a magnetic field.

He was led there by a flaw in Ampere's circuital law, ∮B·dl = μ₀i, which appears when the law is applied just outside a capacitor that is being charged by a time-dependent current i(t).

Take a circular loop of radius r around the wire leading to the capacitor, perpendicular to it and centred on it. By symmetry B runs along the loop with the same size at every point, so the left side is B(2πr) and the law gives B(2πr) = μ₀i(t).

Ampere's law lets us use any surface whose edge is the loop. A flat disc on the loop is pierced by the wire, so the current i crosses it.

Now choose a pot-shaped surface with the same rim, or one shaped like an open tiffin box with a flat bottom, whose bottom lies in the gap between the plates. No charge crosses such a surface anywhere, so the right side becomes zero while the left side is unchanged.

One way B at the point P is non-zero, the other way it is zero. That contradiction means the law was missing a term, one that gives the same B at P whichever surface is used.

The clue is what does cross the surface in the gap: the electric field. With plate area A and charge Q, the field between the plates is E = (Q/A)/ε₀, perpendicular to the plates, uniform over the area A and zero outside it.