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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

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) $C_{t} = \frac{K {鈭�}_{0} A}{d + K d - K a}$ .

(b) $C_{t} = \frac{K C_{0} d}{(d + K d - K a)}$ .

(c) $C_{t} = K C_{0}$ .

Step by step solution

01

Parameters.

$\begin{array}{l} d_{1} = a \\ d_{2} = d - a \\ 鈭� = K {鈭�}_{0} \end{array}$

02

(a) Computing the capacitance.

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

$C_{1} = \frac{鈭� A}{d_{1}}$

And the second of the air is given by

$C_{2} = \frac{鈭� A}{d_{2}}$

The total capacitance is given by

$\frac{1}{C_{t}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{d}{鈭� A} + \frac{d - a}{{鈭�}_{0} A} = \frac{d}{K {鈭�}_{0} A} + \frac{d - a}{{鈭�}_{0} A} = \frac{1}{{鈭�}_{0} A} (\frac{d}{K} + \frac{K d - K a}{K}) = \frac{1}{{鈭�}_{0} A} (\frac{d + K d - K a}{K})$

So the capacitance is given by

$C_{t} = \frac{K {鈭�}_{0} A}{d + K d - K a}$ .

03

(b) Expressing the capacitance as a multiple capacitance.

The capacitance expressed in terms of $C_{0}$ is given by

$C_{t} = \frac{K {鈭�}_{0} A}{d + K d - K a} = \frac{K {鈭�}_{0} A}{(d + K d - K a)} 脳 \frac{d}{d} = \frac{K {鈭�}_{0} A d}{d (d + K d - K a)} = \frac{K C_{0} d}{(d + K d - K a)}$

04

(c)Discussing the capacitance in the limits.

The limit when $a 鈫� 0$ or $a 鈮� 0$ gives $C_{1} = 0$ and the total capacitance is

$\frac{1}{C_{t}} = \frac{1}{C_{2}} = \frac{d}{{鈭�}_{0} A}$

$C_{t} = C_{0}$ (1)

The capacitance when $a 鈫� d$ is given by

$C_{t} = \frac{K C d}{d + K d - K d} = K C_{0}$ .

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A particle of mass 0.195 g carries a charge of $- 2.50 脳 10^{- 8} C$ . The particle is given an initial horizontal velocity that is due north and has magnitude $4.00 脳 10^{4} m / s$ . What are the magnitude and direction of the minimum magnetic field that will keepthe particle moving in the earth鈥檚 gravitational field in the samehorizontal, northward direction?

A rule of thumb used to determine the internal resistance of a source is that it is the open circuit voltage divide by the short circuit current. Is this correct? Why or why not?

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In the circuit of Fig. E25.30, the 5.0 鈩� resistor is removed and replaced by a resistor of unknown resistance R. When this is done, an ideal voltmeter connected across the points band creads 1.9 V. Find (a) the current in the circuit and (b) the resistance R. (c) Graph the potential rises and drops in this circuit (see Fig. 25.20).

When a resistor with resistance Ris connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that Rremains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

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