Problem — Thought experiment, epsilon0 doubles

Problem

Suppose the vacuum permittivity $\varepsilon_0$ suddenly doubles, while $\mu_0$ stays unchanged. How would the following change:

• (a) the capacitance of a parallel-plate capacitor;
• (b) the speed of light in vacuum;
• (c) the energy stored in the capacitor charged to the same voltage?

Solution

(a) Capacitance. For a parallel-plate capacitor $C=\frac{\varepsilon_0 A}{d}$ $C$ is directly proportional to $\varepsilon_0$: if $\varepsilon_0$ doubles, then $\ev{C\ \text{doubles}}$.

(b) Speed of light. From Maxwell’s relation $c=\frac{1}{\sqrt{\varepsilon_0\mu_0}}$ with $\mu_0$ constant, $c\propto 1/\sqrt{\varepsilon_0}$. Doubling $\varepsilon_0$: $c'=\frac{c}{\sqrt2}\approx 0{,}71\,c$ light would slow down by a factor of $1/\sqrt2$.

(c) Stored energy (at constant voltage). The energy of a capacitor as a function of voltage is $U=\frac{1}{2}C V^2$ At fixed $V$, $U$ is proportional to $C$. Since $C$ doubles: $\ev{U\ \text{doubles}}$

Textbook error $U$ halves". At constant voltage, $U=\tfrac12 CV^2$, so $U\propto C$ and the energy doubles, not halves. The error probably arises from using the formula $U=\tfrac{Q^2}{2C}$ (valid at constant charge, not constant voltage): in that different scenario $U$ would indeed halve. Under the assumption in the problem (same voltage), the correct answer is that $U$ doubles.

The textbook incorrectly states that “

Connections

Topics: Electromagnetic waves Concepts: Capacitance and capacitor · Maxwell’s equations Skills: Limiting-case and thought-experiment analysis