Problem 12 - University Physics with Modern Physics

Question

A cylindrical capacitor has an inner conductor of radius 1.5 \(\mathrm{mm}\) and an outer conductor of radius 3.5 \(\mathrm{mm}\) . The two conductors are separated by vacuum, and the entire capacitor is 2.8 \(\mathrm{m}\) long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 \(\mathrm{mV}\) higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

Step-by-Step Solution

Answer

(a) \( C' \approx 5.56 \times 10^{-11} \mathrm{F/m} \). (b) Positive charge on the inner conductor: \( 5.46 \times 10^{-11} \mathrm{C} \); negative charge on the outer conductor: \( -5.46 \times 10^{-11} \mathrm{C} \).

Step 1: Understanding the Capacitance Formula

For a cylindrical capacitor, the capacitance per unit length \( C' \) is given by the formula \( C' = \frac{2 \pi \varepsilon_0}{\ln(b/a)} \), where \( a \) is the radius of the inner conductor and \( b \) is the radius of the outer conductor. \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \mathrm{F/m} \). Convert the radii from millimeters to meters before substituting into the formula.

Step 2: Converting Units

Convert the given radii from millimeters to meters. The inner radius \( a \) is \( 1.5 \) mm or \( 0.0015 \) m, and the outer radius \( b \) is \( 3.5 \) mm or \( 0.0035 \) m.

Step 3: Calculating Capacitance per Unit Length

Substitute the values into the formula: \( C' = \frac{2 \pi \times 8.85 \times 10^{-12}}{\ln(0.0035/0.0015)} \). Calculate this to find \( C' \). After calculation, \( C' \approx 5.56 \times 10^{-11} \mathrm{F/m} \).

Step 4: Calculating Potential Difference

The problem states the potential difference \( V \) between the conductors is 350 mV, or \( 0.35 \) volts.

Step 5: Determining Charge per Unit Length

The charge \( Q' \) per unit length on the conductors is calculated using \( Q' = C' \cdot V \). Substitute the values: \( Q' = 5.56 \times 10^{-11} \cdot 0.35 \). Calculate \( Q' \) to find that \( Q' \approx 1.95 \times 10^{-11} \mathrm{C/m} \).

Step 6: Calculating Total Charge

Since the capacitor is 2.8 m long, the total charge \( Q \) is \( Q = Q' \times 2.8 \). Multiply: \( Q = 1.95 \times 10^{-11} \times 2.8 \). After calculation, \( Q \approx 5.46 \times 10^{-11} \mathrm{C} \).

Step 7: Analyzing Sign of the Charge

The positive charge will be on the inner conductor and the negative charge on the outer conductor, as defined by the direction of the potential difference.

Key Concepts

Capacitance Calculation

The capacitance of a cylindrical capacitor can be particularly interesting to calculate due to its unique geometric shape. Capacitance, in general, refers to the ability of a system to store electric charge. For a cylindrical capacitor, this is typically measured as capacitance per unit length, denoted as \( C' \). To determine \( C' \), we use the formula: \[ C' = \frac{2 \pi \varepsilon_0}{\ln(b/a)} \] Here, \( a \) and \( b \) are the radii of the inner and outer conductors, respectively. It's important to note that \( \varepsilon_0 \) is the permittivity of free space, a constant valued at approximately \( 8.85 \times 10^{-12} \ \mathrm{F/m} \).

• Always remember to convert radii from millimeters to meters for consistency in units.
• The equation reflects how capacitance is inversely related to the natural logarithm of the size ratio of the conductors.

By plugging in the appropriate values, such as \( a = 0.0015 \ \mathrm{m} \) and \( b = 0.0035 \ \mathrm{m} \), one finds the capacitance per unit length to be approximately \( 5.56 \times 10^{-11} \ \mathrm{F/m} \). This value quantifies how much charge is stored per meter of the capacitor when a certain potential difference is applied.

Electric Potential Difference

Electric potential difference is an essential concept when dealing with capacitors. It defines the work needed to move a charge from one conductor to another, and it strongly affects the charges stored. For our cylindrical capacitor, the potential difference \( V \) is given as 350 millivolts, or \( 0.35 \) volts. This value indicates that the inner conductor is positively charged relative to the outer conductor. It's important to understand that the potential difference is proportional to both the charge stored and the capacitance. This relationship can be expressed by the equation: \[ V = \frac{Q}{C} \] In this formula, \( Q \) represents the charge on one conductor, and \( C \) is the total capacitance. It's the potential difference that provides the force to move electrons and build up charge, making it critical in understanding the behavior of capacitors.

• A higher potential difference results in more charge stored if the capacitance remains constant.
• The potential difference helps determine the direction of charge flow between the conductors.

The potential difference paths help ensure the efficient operation of components relying on capacitors in electronic circuits.

Charge Distribution

The distribution of electric charge within a cylindrical capacitor is determined by the potential difference applied across the conductors. When a potential difference is established, as in our example, with the inner conductor being 350 millivolts above the outer one, charges distribute accordingly. Using the relation \( Q' = C' \cdot V \), where \( C' \) is the capacitance per unit length, and \( V \) is the potential difference, we can find the charge per unit length. For this exercise, substituting the respective values gives: \[ Q' = 5.56 \times 10^{-11} \cdot 0.35 \approx 1.95 \times 10^{-11} \ \mathrm{C/m} \] Since the capacitor is 2.8 meters long, the total charge \( Q \) is calculated as: \[ Q = Q' \times L = 1.95 \times 10^{-11} \times 2.8 \approx 5.46 \times 10^{-11} \ \mathrm{C} \]

• The inner conductor will have a positive charge of \( 5.46 \times 10^{-11} \ \mathrm{C} \).
• The outer conductor will have an equal magnitude but opposite charge, indicating a negative sign.

This charge distribution is inherent to the function of capacitors where opposite charges on conductors create an electric field facilitating energy storage. Understanding this concept is vital in electronics as it directly influences how capacitors are used in various applications for energy storage and release.