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Chapter 4: Q62P (page 813)

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 鈫�0 and a 鈫�d.

Short Answer

Expert verified

(a) Ct=K鈭�0Ad+Kd-Ka.

(b) Ct=KC0dd+Kd-Ka.

(c) Ct=KC0.

Step by step solution

01

Parameters.

d1=ad2=d-a鈭�=K鈭�0

02

(a) Computing the capacitance.

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

C1=鈭�Ad1

And the second of the air is given by

C2=鈭�Ad2

The total capacitance is given by

1Ct=1C1+1C2=d鈭�A+d-a鈭�0A=dK鈭�0A+d-a鈭�0A=1鈭�0AdK+Kd-KaK=1鈭�0Ad+Kd-KaK

So the capacitance is given by

Ct=K鈭�0Ad+Kd-Ka.

03

(b) Expressing the capacitance as a multiple capacitance.

The capacitance expressed in terms ofC0 is given by

Ct=K鈭�0Ad+Kd-Ka=K鈭�0Ad+Kd-Ka脳dd=K鈭�0Addd+Kd-Ka=KC0dd+Kd-Ka

04

(c)Discussing the capacitance in the limits.

The limit whena鈫�0 ora鈮�0 givesC1=0 and the total capacitance is

1Ct=1C2=d鈭�0A

Ct=C0 (1)

The capacitance when a鈫�d is given by

Ct=KCdd+Kd-Kd=KC0.

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