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

Short Answer

Expert verified
(a) Charge on each conductor is 1.60 × 10^{-9} C. (b) Ratio of radii is approximately 1.0868.

Step by step solution

01

Use Energy Formula for Capacitors

The energy stored in a capacitor is given by the formula \[U = \frac{1}{2} C V^2\]where \( U \) is the energy, \( C \) is the capacitance, and \( V \) is the potential difference. We can rearrange this formula to solve for capacitance \( C \):\[C = \frac{2U}{V^2}\]
02

Calculate Capacitance

Substitute the given values into the formula:\[C = \frac{2 \times 3.20 \times 10^{-9}}{4.00^2}\]Calculate C:\[C = \frac{6.40 \times 10^{-9}}{16} = 4.00 \times 10^{-10} \text{ F}\]
03

Use Capacitance Formula to Find Charge

The charge \( Q \) on the conductors is given by:\[Q = CV\]Substituting the known values:\[Q = (4.00 \times 10^{-10} \text{ F})(4.00 \text{ V})\]Calculate \( Q \):\[Q = 1.60 \times 10^{-9} \text{ C}\]
04

Use Capacitance Formula for Cylinders

The capacitance \( C \) of a cylindrical capacitor is given by:\[C = \frac{2\pi \varepsilon_0 L}{\ln\left(\frac{b}{a}\right)}\]where \( L \) is the length, \( a \) is the radius of the inner cylinder, \( b \) is the radius of the outer cylinder, and \( \varepsilon_0 \) is the vacuum permittivity \((8.85 \times 10^{-12} \text{ F/m})\).
05

Solve for the Ratio of Radii

From the given formula:\[\frac{b}{a} = e^{\left(\frac{2\pi \varepsilon_0 L}{C}\right)}\]Substitute the values for calculation:\[\frac{b}{a} = e^{\left(\frac{2\pi (8.85 \times 10^{-12}) (15.0)}{4.00 \times 10^{-10}}\right)}\]Calculate inside the exponent and then solve the exponent:\[\frac{b}{a} = e^{0.083387}\]\[\frac{b}{a} \approx 1.0868\]
06

Conclusion

With the charge calculated as \(1.60 \times 10^{-9} \text{ C}\) and the ratio of radii approximately \( 1.0868 \), the exercise is complete.

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Key Concepts

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

Capacitance
Capacitance is a fundamental property of a capacitor. It reflects a capacitor's ability to store an electric charge per unit of electrical potential difference across its conductors. Mathematically, it is represented as \( C = \frac{Q}{V} \),where \( C \) is the capacitance in farads (F), \( Q \) is the charge in coulombs (C), and \( V \) is the potential difference in volts (V).
For different shapes and materials, capacitors have unique formulas for calculating capacitance, but the principle remains the same: it measures how effectively a capacitor can store electrical energy given a certain voltage. In cylindrical capacitors, the capacitance also depends on their physical dimensions, such as length, as well as the radii of the conducting surfaces.
Understanding capacitance is essential because it helps predict how much energy a capacitor can store, making it crucial for designing electrical circuits and systems.
Charge Calculation
Calculating the charge on a capacitor involves understanding the relationship between charge, capacitance, and voltage. The charge \( Q \) stored in a capacitor with a capacitance \( C \) and a voltage \( V \) applied across it is given by:\[ Q = C \times V \]This formula implies that the charge stored is directly proportional to both the capacitance and the voltage.
In the context of our problem, with a given capacitance of \( 4.00 \times 10^{-10} \) F (farads) and a potential difference of 4.00 V, calculating the charge requires just simple multiplication:\[ Q = (4.00 \times 10^{-10} \, \text{F})(4.00 \, \text{V}) = 1.60 \times 10^{-9} \, \text{C} \]This equation highlights the key mechanism of energy storage in capacitors by means of charge accumulation between its plates. It’s a straightforward but powerful concept that underscores much of capacitor functionality in electronics.
Potential Difference
The potential difference, or voltage, across a capacitor is a critical factor that determines the amount of charge and energy it can store. It represents the work needed to move a unit charge between two points in an electric field.
In simple terms, potential difference is what pushes electrons through a circuit and into the capacitor’s plates. It's calculated as:\[ V = \frac{U}{Q} \],connecting energy \( U \) and charge \( Q \).
In our example, you are given a potential difference of 4.00 V. This value is essential for calculating both the charge on the capacitor and ultimately the energy stored. To maintain safety and efficiency, especially in practical applications, it’s important that capacitors are not subjected to voltages higher than their rated capacity to prevent the risk of damage or failure.
Energy Stored in Capacitor
The energy stored in a capacitor is determined by the formula:\[ U = \frac{1}{2} C V^2 \]This equation shows that the energy \( U \) is dependent on both the capacitance \( C \) and the square of the potential difference \( V \).
This relationship demonstrates how even a small increase in voltage can significantly increase the energy stored, making voltage levels critical in capacitor design.
In this exercise, the energy already stored is 3.20 \( \times \) 10\(^{-9}\) J. With this information, we can confirm the capacitor's effectiveness by ensuring that the energy calculation aligns with expected performance parameters. It’s crucial for applications in power supply systems, where accurate energy storage predictions contribute to stable and reliable power delivery.
Vacuum Permittivity
Vacuum permittivity, denoted as \( \varepsilon_0 \), is a constant value representing the ability of a vacuum to permit electric field lines. It is a critical factor in calculating the capacitance of various capacitor configurations.
For a cylindrical capacitor, the capacitance is given by:\[ C = \frac{2\pi \varepsilon_0 L}{\ln\left(\frac{b}{a}\right)} \],where \( L \) is the length, \( a \) is the radius of the inner cylinder, and \( b \) is the radius of the outer cylinder.
\( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \text{ F/m} \) (farads per meter). This constant is fundamental in the laws that govern electromagnetism, describing how electric fields interact in free space.
By influencing capacitance, vacuum permittivity affects how capacitors are designed and how they function in electrical circuits, playing a decisive role in innovations within the electronics industry.

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Most popular questions from this chapter

You have two identical capacitors and an external potential source. (a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel. (b) Compare the maximum amount of charge stored in each case. (c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E = 3.20\times10^5\space V/m\). When the space is filled with dielectric, the electric field is \(E = 2.50\times10^5\space V/m\). (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

Your electronics company has several identical capacitors with capacitance \(C_1\) and several others with capacitance \(C_2\). You must determine the values of \(C_1\) and \(C_2\) but don't have access to \(C_1\) and \(C_2\) individually. Instead, you have a network with \(C_1\) and \(C_2\) connected in series and a network with \(C_1\) and \(C_2\) connected in parallel. You have a 200.0-V battery and instrumentation that measures the total energy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, 0.180 J of energy is stored in the network. When the series combination is connected, 0.0400 J of energy is stored. You are told that \(C_1\) is greater than \(C_2\). (a) Calculate \(C_1\) and \(C_2\). (b) For the series combination, does \(C_1\) or \(C_2\) store more charge, or are the values equal? Does \(C_1\) or \(C_2\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.5 cm, and the outer sphere has radius 14.8 cm. A potential difference of 120 V is applied to the capacitor. (a) What is the energy density at \(r\) = 12.6 cm, just outside the inner sphere? (b) What is the energy density at \(r\) = 14.7 cm, just inside the outer sphere? (c) For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

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 (\(\textbf{Fig. P24.62}\)). (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 \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

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