Electromagnetism · Experiment

Capacitor & Dielectrics

Two parallel plates store charge and energy in the field between them. Resize the plates, stretch the gap, then slide a dielectric slab in — and see why what happens depends on whether the battery is still attached.

E field (uniform, E = V/d)dielectric slab κplates & capacitance bar± polarization charges on the slab faces

Controls

Capacitance C—
Charge Q = CV—
Voltage V—
Field E = V/d—
Energy U = ½CV²—
Battery connected: V is pinned. C = ε₀A/d — grow the plates or shrink the gap and the capacitor holds more charge at the same voltage.
About this experiment ›

What you are looking at

A parallel-plate capacitor seen edge-on: the violet plates carry + and − charge, the gold arrows show the electric field between them, and the green slab is a dielectric you can slide into the gap. The bars on the right track the capacitance and stored energy live.

Capacitance is pure geometry

A capacitor's ability to hold charge per volt depends only on its shape — big plates close together store more:
C = ε₀·A / d
With a dielectric filling the gap, the slab's molecules polarize — their charges shift slightly and build a surface charge that opposes the field. The same plate charge now produces less voltage, which means more capacitance:
C = κ·ε₀·A / d
A partially-inserted slab acts as two capacitors side by side (filled and empty), added in parallel — so C climbs smoothly from ε₀A/d to κε₀A/d as you push it in. Because the plates are conductors, the gap field stays uniform at E = V/d everywhere.

The experiment that matters: battery vs isolated

What happens when C changes depends on what is held fixed:
Battery connected: V fixed → Q = CV rises, E = V/d unchanged, U = ½CV² rises
Isolated: Q fixed → V = Q/C falls, E falls by κ, U = Q²/2C falls
The isolated case is the striking one: push the slab in and the voltmeter reading actually drops — the trapped charge hasn't changed, but the polarized dielectric partially cancels its field.

Why the slab gets pulled in

In the isolated case the stored energy U = Q²/2C strictly decreases as the slab enters — and systems feel a force toward lower energy, so the fringing field grips the dielectric and drags it inward. With the battery attached U rises, but the battery supplies exactly twice the increase, so the net energy still drops and the pull is inward either way. This tiny tug is real — it's how some microphones and MEMS sensors work.

Things to try

Charge up, disconnect the battery, then slide the slab in and watch V plunge by κ while Q stays frozen. Reconnect the battery and repeat: now V holds and Q climbs instead. Halve the gap d and watch C double — then check that E = V/d doubles too (battery mode) since the same volts now span half the distance.