Physics

Problem

Four parallel-plate capacitors, each having the same plate separation

d

, are listed below. In every case the space between the plates is completely filled by the stated medium. C

_1

: air, plate area

A

_2

: air, plate area

2A

_3

: glass (

\kappa = 4

), plate area

A

_4

: mica (

\kappa = 6

), plate area

\tfrac{A}{2}

(a) Rank the four capacitors in descending order of capacitance. Justify each step of your ranking without performing any numerical calculation. (b) If each capacitor is separately charged to the same potential difference

V

and then disconnected from the battery, rank them in descending order of the electrostatic energy stored. Give reasons for your answer. (c) If instead each capacitor is connected across separate identical batteries that maintain the potential difference

V

while charging and afterwards, comment on whether your ranking in part (b) will change. Support your answer with qualitative reasoning.

Intuitive explanation

Part (a): Ranking by Capacitance

Capacitance depends on three factors: plate area (larger is better), plate separation (smaller is better), and the dielectric constant $\kappa$ of the medium (higher is better). The formula $C = \kappa \epsilon_0 A / d$ shows that capacitance scales linearly with $\kappa$ and $A$ , but inversely with $d$ . Since all four capacitors share the same $d$ , you need only compare the product $\kappa A$ for each: $C_1$ has $1 \cdot A$ , $C_2$ has $1 \cdot 2A$ , $C_3$ has $4 \cdot A$ , and $C_4$ has $6 \cdot \frac{A}{2}$ . Rank them by these products to find the order.

Part (b): Ranking by Energy (Disconnected)

When a capacitor is charged to voltage $V$ and then disconnected, the charge $Q$ on it becomes fixed. The stored energy is $U = \frac{

Intuitive explanation

Part (a): Ranking by Capacitance

Part (b): Ranking by Energy (Disconnected)

When a capacitor is charged to voltage $V$ and then disconnected, the charge $Q$ on it becomes fixed. The stored energy is $U = \frac{

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