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 (Fig. P24.62). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C0 when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits a $\mathbf{\to }$0 and a $\mathbf{\to }$d.

Short answer

(a) $C_t=\frac{K{\in }_{0}A}{d+Kd-Ka}$.

(b) $C_t=\frac{K{C}_{0}d}{\left(d+Kd-Ka\right)}$.

(c) $C_t=K{C}_0$.

Step-by-step solution

Parameters.

$\begin{array}{l}{d}_1=a\\ d_2=d-a\\ \in =K{\in }_0\end{array}$

(a) Computing the capacitance.

Consider the two capacitors in series as each capacitance is given by

$C_1=\frac{\in A}{d_1}$

And the second of the air is given by

$C_2=\frac{\in A}{d_2}$

The total capacitance is given by

$$\begin{gathered} \frac{1}{C_t}=\frac{1}{C_1}+\frac{1}{C_2} \\ =\frac{d}{\in A}+\frac{d-a}{{\in }_{0}A} \\ =\frac{d}{K{\in }_{0}A}+\frac{d-a}{{\in }_{0}A} \\ =\frac{1}{{\in }_{0}A}\left(\frac{d}{K}+\frac{Kd-Ka}{K}\right) \\ =\frac{1}{{\in }_{0}A}\left(\frac{d+Kd-Ka}{K}\right) \end{gathered}$$

So the capacitance is given by

$C_t=\frac{K{\in }_{0}A}{d+Kd-Ka}$.

(b) Expressing the capacitance as a multiple capacitance.

The capacitance expressed in terms of$C_0$ is given by

$$\begin{gathered} C_t=\frac{K{\in }_{0}A}{d+Kd-Ka} \\ =\frac{K{\in }_{0}A}{\left(d+Kd-Ka\right)}\times \frac{d}{d} \\ =\frac{K{\in }_{0}Ad}{d\left(d+Kd-Ka\right)} \\ =\frac{K{C}_{0}d}{\left(d+Kd-Ka\right)} \end{gathered}$$

(c)Discussing the capacitance in the limits.

The limit when$a\to 0$ or$a\approx 0$ gives$C_1=0$ and the total capacitance is

$\frac{1}{C_t}=\frac{1}{C_2}=\frac{d}{{\in }_{0}A}$

$C_t=C_0$ (1)

The capacitance when $a\to d$ is given by

$C_t=\frac{KCd}{d+Kd-Kd}=K{C}_0$.