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24.77. Three square metal plates \(A, B,\) and \(C,\) each 12.0 \(\mathrm{cm}\) on a side and 1.50 \(\mathrm{mm}\) thick, are arranged as in Fig. 24.43 . The plates are separated by sheets of paper 0.45 \(\mathrm{mm}\) thick and with dielectric constant \(4.2 .\) The outer plates are connected together and connected to point \(b\) . The inner plate is connected to point \(a\) (a) Copy the diagram and show by plus and minus signs the charge distribution on the plates when point \(a\) is maintained at a positive potential relative to point \(b\) . (b) What is the capacitance between points \(a\) and \(b ?\)

Short Answer

Expert verified
The capacitance between points \( a \) and \( b \) is 5.95 pF.

Step by step solution

01

Understanding the Plate Arrangement

There are three square metal plates: A, B, and C. Plates A and C are the outer plates, connected together and to point \( b \), while plate B is the inner plate connected to point \( a \). Each plate is a square with 12.0 cm sides and 1.50 mm thickness. The plates are separated by sheets of paper, each 0.45 mm thick with a dielectric constant of 4.2.
02

Analyze Charge Distribution

When point \( a \) is at a higher potential than point \( b \), the inner plate (B) connected to point \( a \) will acquire a positive charge. The outer plates (A and C) connected to point \( b \) will have an induced negative charge on the side facing plate B. This opposite charge configuration is due to the applied electric field between the plates.
03

Identify Capacitor Configuration

The system is equivalent to two capacitors in series: one between plates A and B, and another between plates B and C. The connection of A and C to the same point forms these two capacitors in series across point \( a \) and \( b \).
04

Calculate Capacitance for a Single Capacitor

Use the formula for capacitance of a parallel plate capacitor: \[ C = \varepsilon \frac{A}{d} \]Here, \( \varepsilon = \varepsilon_0 \times \varepsilon_r \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m is the permittivity of free space, and \( \varepsilon_r = 4.2 \) is the dielectric constant.The area \( A = (0.12 \text{ m})^2 = 0.0144 \text{ m}^2 \), and the distance between plates is \( d = 0.45 \text{ mm} = 0.00045 \text{ m} \).
05

Substitute Values into the Capacitance Formula

Calculate \( \varepsilon = 8.85 \times 10^{-12} \times 4.2 = 3.717 \times 10^{-11} \) F/m.Substitute into the formula for capacitance: \[ C = 3.717 \times 10^{-11} \times \frac{0.0144}{0.00045} \approx 1.19 \times 10^{-11} \text{ F} = 11.9 \text{ pF} \].
06

Calculate Total Capacitance for Series Capacitors

For two capacitors in series, the total capacitance \( C_t \) is given by:\[ \frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} \]Since both capacitors have the same capacitance (11.9 pF), this simplifies to:\[ \frac{1}{C_t} = \frac{1}{11.9} + \frac{1}{11.9} = \frac{2}{11.9} \].
07

Solve for Total Capacitance

Calculate \( C_t \):\[ C_t = \frac{11.9}{2} = 5.95 \text{ pF} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitor in Series
When we talk about capacitors in series, we are referring to a specific arrangement of capacitors in a circuit. In this setup, the capacitors are connected end to end, forming a chain. This means the total voltage across the series is the sum of the voltages across each individual capacitor. However, the charge stored in each capacitor is the same.
When working with capacitors in series, one of the main characteristics is how to calculate the total or equivalent capacitance, denoted as \( C_t \). You can find \( C_t \) using the formula: \[ \frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \] where \( C_1, C_2, \ldots \) are the capacitances of the individual capacitors.
This formula shows that the total capacitance is less than any individual capacitor's capacitance because capacitors in series divide the capacitance.
Dielectric Constant
The dielectric constant, often represented by \( \varepsilon_r \), is a measure of a material's ability to increase the capacitance of a capacitor when the material fills the space between the plates. Essentially, it is a factor that indicates how much the dielectric can reduce the electric field, thus increasing the stored energy.
In the given problem, paper is used as the dielectric material with a dielectric constant of 4.2. This means that when the paper is placed between the plates, the capacitance becomes 4.2 times greater than it would be if the space were just a vacuum.
The relationship between dielectric constant and capacitance is important because it explains how different materials can drastically change a capacitor's behavior and efficiency in a circuit.
Electric Field
The concept of an electric field is crucial in understanding how capacitors work. It represents the force that a charge exerts in its surrounding space and is generated between the plates of a capacitor when there is a voltage difference.
In a simple parallel plate capacitor, the electric field \( E \) between the plates can be calculated using the formula \( E = \frac{V}{d} \), where \( V \) is the voltage across the plates and \( d \) is the separation between them.
The electric field is uniform between the plates, and its direction is from the positive plate to the negative. This field plays a significant role in determining the amount of charge the capacitor can hold, as well as the energy stored within it.
Permittivity
Permittivity is a fundamental concept related to dielectrics in capacitors. It signifies how an electric field affects, and is affected by, a dielectric medium. Permittivity is often symbolized by \( \varepsilon \) and is composed of two parts: the permittivity of free space \( \varepsilon_0 \) and the relative permittivity \( \varepsilon_r \).
The permittivity of free space, \( \varepsilon_0 \), is a constant approximately equal to \( 8.85 \times 10^{-12} \) F/m, while \( \varepsilon_r \) is the dielectric constant of the material. Together, they determine the effective permittivity \( \varepsilon = \varepsilon_0 \times \varepsilon_r \).
This effective permittivity plays a crucial role in determining the capacitance \( C \) of a capacitor through the equation \( C = \varepsilon \frac{A}{d} \), where \( A \) is the area of the plates and \( d \) is the distance between them. Understanding permittivity helps us understand how different materials affect capacitance and the storage of electric energy.

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Most popular questions from this chapter

24.46. A parallel-plate capacitor has capacitance \(C=12.5 \mathrm{pF}\) when the volume between the plates is filled with air. The plates are circular, with radius 3.00 \(\mathrm{cm}\) . The capacitor is connected to a battery and a charge of magnitude 25.0 \(\mathrm{pC}\) goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 \(\mathrm{pC}\) (a) What is the dielectric constant \(K\) of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been insered? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

24.8. A \(5.00-\) pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to \(1.00 \times 10^{2} \mathrm{V}\) . The electric field between the plates is to be no greater than \(1.00 \times 10^{4} \mathrm{N} / \mathrm{C}\) . As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

24.26. An air capacitor is made from two flat parallel plates 1.50 \(\mathrm{mm}\) apart. The magnitude of charge on each plate is 0.0180\(\mu \mathrm{C}\) when the potential difference is 200 \(\mathrm{V}\) . (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m}\) . (d) When the charge is 0.0180\(\mu \mathrm{C}\) , what total energy is stored?

24.61. Three capacitors having capacitances of \(8.4,8.4,\) and 4.2\(\mu \mathrm{F}\) are connected in series across a \(36-\mathrm{V}\) potential difference. (a) What is the charge on the \(4.2-\mu \mathrm{F}\) capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

24.35. A cylindrical air capacitor of length 15.0 m stores \(3.20 \times 10^{-9} \mathrm{J}\) of energy when the potential difference between the two conductors is 4.00 \(\mathrm{V}\) . (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

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