Ranking the energy of capacitors

Three parallel-plate capacitors, identical in plate dimensions but with different separations and dielectrics, are all charged to the same voltage $V$:

• A: separation $d$, vacuum;
• B: separation $2d$, vacuum;
• C: separation $d$, dielectric $\varepsilon_r = 2$.

Rank the stored energy from smallest to largest.

Solution

Setup. At fixed voltage it is convenient to use the energy in the form $U = \tfrac{1}{2}\,C\,V^2$ because $V$ is the same for all: the energy then depends only on the capacitance. For a parallel-plate capacitor $C = \varepsilon_r\,\varepsilon_0\,\frac{A}{d}$ with $A$ the same for all. Comparing the capacitances. $\begin{aligned} C_A &= \varepsilon_0\frac{A}{d} \\ C_B &= \varepsilon_0\frac{A}{2d} = \tfrac{1}{2}C_A \\ C_C &= 2\,\varepsilon_0\frac{A}{d} = 2\,C_A \end{aligned}$ Ranking. Since $U \propto C$ at constant $V$, the energy ranking follows the capacitance ranking $C_B < C_A < C_C$: $\ev{U_B < U_A < U_C}$ Physical reading. Doubling the separation (B) halves the capacitance and hence the energy; inserting a dielectric $\varepsilon_r = 2$ (C) doubles it, because at fixed $V$ the dielectric allows more charge to accumulate.

Links

Topics: Electric field and potential Concepts: Capacitance and capacitor · Dielectrics and polarisation · Field energy density Skills: Superposition principle Methods: Gradient method Objects: Parallel-plate capacitor