Question

An air capacitor is made by using two flat plates, each with area A, separated by a distance $d$. Then a metal slab having thickness $a$ (less than $d$) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate ($\mathbf{\text{Fig. P24.62}}$). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C${}_0$ when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits $a$ $\to$ 0 and $a$ $\to$ d.

Short answer

(a) Total capacitance is $C=\frac{{\epsilon }_{0}A}{d-a}$. (b) Capacitance as multiple of $C_0$ is $\frac{C_0}{1-\frac{a}{d}}$. (c) As $a\to 0$, $C\to C_0$; as $a\to d$, $C\to \mathrm{\infty }$.

Step-by-step solution

Understanding the Problem

We need to find the capacitance of a capacitor with two parallel plates separated by a distance $d$ and having a metal slab of thickness $a$ inserted between them. The challenge is to figure out how the capacitance changes with this insertion.

Concept of Series Capacitance

The metal slab divides the capacitor into two capacitors connected in series: one with separation $d-a$ and one with slab thickness $a$. We need to calculate the individual capacitances of these two parts and combine them to find the total capacitance.

Calculate Capacitance without Metal

The capacitance between the two plates without the metal sheet is given by $C_0=\frac{{\epsilon }_{0}A}{d}$, where ${\epsilon }_0$ is the permittivity of free space and $A$ is the area of the plates.

Capacitance of Each Section

The system can be divided into two capacitors: the first from one plate to one side of the slab with distance $d-a$ and the second through the thickness of the slab. However, in this scenario, with the slab being conductive, it doesn't introduce a separate capacitance for slab thickness. The first section's capacitance is $C_1=\frac{{\epsilon }_{0}A}{d-a}$ and the section from slab to the second plate is $C_2=\frac{{\epsilon }_{0}A}{d-a}$.

Total Capacitance in Series

For capacitors in series, the total capacitance $C$ is found via $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$. Using symmetry for $C_1$ and $C_2$, this simplifies to $C=\frac{2C_1{C}_2}{C_1+C_2}=\frac{{\epsilon }_{0}A}{d-a}$.

Capacitance Ratio with $C_0$

Express the capacitance with the metal slab $C$ as a multiple of $C_0$: $C=\frac{d}{d-a}{C}_0$.

Limit Analysis for Slab Thickness

As $a$ approaches 0, the metal slab has negligible effect, and $C\to C_0$. As $a$ approaches $d$, the central slab dominates, resulting in an increase in capacitance, theoretically becoming infinite as the effective gap approaches zero.

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor is a simple yet powerful device that consists of two conductive plates facing each other, separated by a small distance. The plates are usually flat and are designed to hold opposite electric charges, creating an electric field between them. This field stores electrical energy.
The capacitance of a parallel plate capacitor can be calculated using the formula: $$C_0=\frac{{\epsilon }_{0}A}{d}$$,where:

• $C_0$ is the capacitance,
• ${\epsilon }_0$ is the permittivity of free space,
• $A$ is the area of one of the plates, and
• $d$ is the distance between the plates.

Understanding the structure and function of a parallel plate capacitor is fundamental to grasp more complex capacitor systems, such as when modifications such as inserting a metal slab are involved.

Metal Slab Insertion

When a metal slab is inserted between the plates of a parallel plate capacitor, it significantly alters the capacitance. This slab has a thickness $a$, which is less than the separation distance $d$ between the plates. Even though the slab doesn’t touch the plates, it divides the space into two regions, essentially forming two capacitors in series.The metal slab increases the capacitance because it shortens the effective distance between oppositely charged regions, enhancing the electric field strength. This alteration can be visualized as two new capacitors, each having a separation of $d-a$.
Calculating the modified capacitance involves understanding these regions:

• The effective separation between the two outside faces of the slab itself.
• The reduced effective distance $d-a$.

The capacitance increases approximately as a factor of $\frac{d}{d-a}$ times $C_0$, as derived from analyzing capacitors in series.

Series Capacitance

Capacitors in series add up differently than those in parallel. When we consider two capacitors $C_1$ and $C_2$ in series, the total or equivalent capacitance $C$ can be calculated with:$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$$This results in a capacitance lower than that of the smallest individual capacitor. The metal slab's insertion modifies the capacitor into two sections, both having a capacitance $$C_1=C_2=\frac{{\epsilon }_{0}A}{d-a}$$Considering these capacitances as in series, the combined capacitance $C$ for the entire system becomes:$$C=\frac{2C_1{C}_2}{C_1+C_2}=\frac{{\epsilon }_{0}A}{d-a}$$.This analysis shows how the slab effectively reduces the gap between opposing charges, resulting in enhanced capacitance. This knowledge is particularly vital when assessing changes like metal slab insertions or thicker dielectric materials.

Permittivity of Free Space

The permittivity of free space, often denoted ${\epsilon }_0$, is a constant that plays a critical role in the calculation of capacitance. It is the measure of how much capacitance is allowed for a capacitor placed in a vacuum or air. Its value is approximately $8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}$.In a parallel plate capacitor, the permittivity of free space is a key factor because it affects how much charge can be stored per unit voltage. Together with the plate area and the separation distance, it determines the basic capacitance of the capacitor prior to any insertions or modifications like a metal slab.Understanding ${\epsilon }_0$ is essential when calculating capacitance in various scenarios, as it allows us to compare different capacitor configurations and the effects of dielectric materials or conductive inserts, like the metal slab, on total capacitance.