Energy and dielectric: battery connected or disconnected

The energy stored in a capacitor is always

$U = \tfrac12 C V^2 = \frac{Q^2}{2C}$

But what happens to the energy when we insert a dielectric depends crucially on how we are running the experiment — that is, whether the battery is still connected or not. The two situations lead to opposite conclusions, and this is an excellent test bench for truly understanding the formulas.

Note

Constant charge (battery disconnected). Inserting the dielectric increases $C$ by a factor $\varepsilon_r$; but $Q$ is fixed (no wire for it to enter or leave through), so $V = Q/C$ falls by $\varepsilon_r$. The energy $U = Q^2/(2C)$ decreases by $\varepsilon_r$. Where does the missing energy go? It is the work done by the dielectric, literally “sucked” into the capacitor by the electrostatic forces: there is a net attraction that pulls in the partially inserted slab.

Note

Constant voltage (battery connected). Now $V$ is fixed by the battery; $C$ grows by $\varepsilon_r$ and so $Q = CV$ increases by $\varepsilon_r$ — the battery supplies additional charge. The energy $U = \tfrac12 C V^2$ increases by $\varepsilon_r$. The battery actually does twice the work: half ends up in the capacitor, half in mechanical work on the dielectric and dissipation in the wires.

Principle — Summary

Same piece of dielectric, two opposite situations: with $Q$ constant the energy falls, with $V$ constant the energy rises. The “trick” of the battery supplying or absorbing charge makes all the difference.

Example — Partially filled capacitor

A parallel-plate capacitor has plates of area $A = 100\;\text{cm}^2 = 10^{-2}\;\text{m}^2$ separated by $D = 1{,}0$ mm. Half the thickness is occupied by a dielectric with $\varepsilon_r = 4$ (for example mica), the other half is air. Calculate the total capacitance.

The structure is equivalent to two capacitors in series, one of thickness $D/2$ with $\varepsilon_r = 1$ and one of thickness $D/2$ with $\varepsilon_r = 4$: $C_1 = \frac{\varepsilon_0 A}{D/2} = 2C_0, \qquad C_2 = \frac{4\varepsilon_0 A}{D/2} = 8C_0$ with $C_0 = \varepsilon_0 A / D \approx 8{,}85\;\text{pF}$. In series: $C_\text{tot} = \frac{C_1 C_2}{C_1 + C_2} = \frac{16 C_0}{10} = 1{,}6\,C_0 \approx 14{,}2\;\text{pF}$ Comparison: a single dielectric with $\varepsilon_r = 4$ filling the whole space would have given $4 C_0 \approx 35{,}4$ pF. The fact that half is still air heavily penalises the capacitance: in the series combination, it is the low-$\varepsilon_r$ region that dominates the result.

Curiosity — Dielectric strength and maximum voltage

Under a sufficiently intense field (the dielectric strength) any dielectric “gives way”: electrons are torn from the nuclei and the material becomes conducting — a laboratory-scale lightning bolt. For air the strength is $\sim 3$ MV/m; for mica $\sim 100$ MV/m. This is why a real capacitor always has two specifications: capacitance and maximum voltage. Exceeding the latter means destroying it.

Links

Topics: Electric field and potential Concepts: Capacitance and capacitor · Dielectrics and polarisation · Field energy density

Related exercises: Ranking the energy of capacitors · Field energy in the capacitor · Energy of a capacitor