Question

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

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

Step-by-step solution

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 \( C \):\[C = \frac{2U}{V^2}\]

Calculate Capacitance

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

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 \( Q \):\[Q = 1.60 \times 10^{-9} \text{ C}\]

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 \((8.85 \times 10^{-12} \text{ F/m})\).

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:\[\frac{b}{a} = e^{0.083387}\]\[\frac{b}{a} \approx 1.0868\]

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.

Key concepts

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.