Ranking capacitors with different geometries

Problem

Ranking: capacitors with different geometries. Four capacitors, all with the same plate area $A = 1\;\text{m}^2$ but different geometries: (A) parallel-plate, $d = 0{,}5\;\text{mm}$; (B) parallel-plate with dielectric $\kappa = 4$, $d = 0{,}5\;\text{mm}$; (C) parallel-plate, $d = 2\;\text{mm}$; (D) parallel-plate with $\kappa = 2$, $d = 1\;\text{mm}$. Order them from greatest to smallest capacitance. Hint: compare the ratios $\kappa/d$.

Solution

Reference formula. For a parallel-plate capacitor with a dielectric: $C = \frac{\kappa\varepsilon_0 A}{d}.$ Since $\varepsilon_0$ and $A$ are the same for all of them, the capacitance is proportional only to the ratio $\kappa/d$: we just need to compare that (with $d$ in mm).

Comparing the ratios $\kappa/d$. $A:\ \frac{1}{0{,}5} = 2, \qquad B:\ \frac{4}{0{,}5} = 8, \qquad C:\ \frac{1}{2} = 0{,}5, \qquad D:\ \frac{2}{1} = 2.$

Ordering. In decreasing order of $\kappa/d$: $\ev{B > A = D > C}.$ B’s dielectric ($\kappa=4$) puts it at the top; A and D tie (D has double the thickness but also double $\kappa$, and the two effects exactly cancel out); C, the thickest with no dielectric, is the worst. The moral: to increase capacitance you can either reduce the distance between the plates or insert a dielectric with a high $\kappa$ — what matters is the ratio between the two.

Links

Topics: Campo elettrico e potenziale Concepts: Capacità e condensatore · Dielettrici e polarizzazione Methods: Semplificazione serie-parallelo Objects: Condensatore piano