Problem 31 - University Physics with Modern Physics

Question

A very long wire carries a uniform linear charge density \(\lambda\). Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed 2.50 \(\mathrm{cm}\) from the wire and the other probe is 1.00 \(\mathrm{cm}\) farther from the wire, the meter reads 575 \(\mathrm{V}\) (a) What is \(\lambda ?\) (b) If you now place one probe at 3.50 \(\mathrm{cm}\) from the wire and the other probe 1.00 \(\mathrm{cm}\) farther away, will the voltmeter read 575 \(\mathrm{V}\) ? If not, will it read more or less than 575 \(\mathrm{V}\) ? Why? (c) If you place both probes 3.50 \(\mathrm{cm}\) from the wire but 17.0 \(\mathrm{cm}\) from each other, what will the voltmeter read?

Step-by-Step Solution

Answer

(a) \( \lambda \approx 2.38 \times 10^{-9} \ \mathrm{C/m} \), (b) less than 575 V, (c) 0 V.

Step 1: Understanding Potential Difference

The potential difference between two points near a long charged line can be given by the equation: \[ V = \frac{\lambda}{2\pi \varepsilon_0} \ln \left( \frac{r_2}{r_1} \right) \]where \( \lambda \) is the linear charge density, \( r_1 \) and \( r_2 \) are the distances from the wire to the points, and \( \varepsilon_0 \) is the vacuum permittivity \( (8.85 \times 10^{-12} \ \mathrm{C^2/N \, . \, m^2}) \).

Step 2: Solving for λ

We are given that the voltmeter reads 575 V when one probe is at 2.50 cm (\(0.025 \ \mathrm{m}\)) and the other is at 3.50 cm (\(0.035 \ \mathrm{m}\)). Substituting these into the potential difference formula, we have:\[ 575 = \frac{\lambda}{2\pi \times 8.85 \times 10^{-12}} \ln \left( \frac{0.035}{0.025} \right) \]We need to solve for \( \lambda \).

Step 3: Calculating the Natural Logarithm

Calculate the natural logarithm: \[ \ln \left( \frac{0.035}{0.025} \right) = \ln(1.4) \approx 0.3365 \]

Step 4: Solving for λ

Rearranging the formula from step 2 to solve for \( \lambda \):\[ \lambda = \frac{575 \times 2\pi \times 8.85 \times 10^{-12}}{0.3365} \]Calculating this, we find:\[ \lambda \approx 2.38 \times 10^{-9} \ \mathrm{C/m} \]

Step 5: Analyzing New Probe Placement (Part b)

Now, place one probe at 3.50 cm (0.035 m) and the other probe 1.00 cm farther, or 4.50 cm (0.045 m). Using the same formula with the previous λ value:\[ V = \frac{\lambda}{2\pi \varepsilon_0} \ln \left( \frac{0.045}{0.035} \right) \]Calculate the natural log:\[ \ln \left( \frac{0.045}{0.035} \right) = \ln(1.286) \approx 0.2513 \]Substituting to find the voltage, the voltmeter will not read 575 V; it will be less.

Step 6: Placing Probes 17 cm Apart at Constant Distance (Part c)

If both probes are placed at 3.50 cm but 17.0 cm apart, this means there is no potential difference due to their equal radial distance from the wire. Hence, the wire will read 0 V because the potential depends only on the radial distance from the wire, not the distance between probes.

Key Concepts

Linear Charge Density

Linear charge density is a fundamental concept in electrostatics. It represents how much electric charge is distributed along a line, such as a wire. The symbol for linear charge density is typically \(\lambda\), and it has units of coulombs per meter (C/m).

Linear charge density is crucial when dealing with infinitely long charged wires. Such wires create electric fields that vary with distance from the wire. By knowing \(\lambda\), you can predict how these fields behave, which helps solve problems involving potential differences around charged wires.

Understanding \(\lambda\) allows you to solve equations like the one used in the exercise, where the potential difference between two points is calculated by integrating the electric field produced by the charged wire.

Potential Difference

The potential difference, often referred to as voltage, measures the electric potential between two points. This difference tells you how much work would be done by or against the electric field to move a test charge between those points. Mathematically, for a long charged wire, it can be expressed as: \[ V = \frac{\lambda}{2\pi \varepsilon_0} \ln \left( \frac{r_2}{r_1} \right) \]where \(\lambda\) is the linear charge density, \(r_1\) and \(r_2\) are distances from the wire, and \(\varepsilon_0\) is the vacuum permittivity.

In the exercise, we used this equation to find the potential difference when probes were placed at different distances from the wire. By calculating the logarithm of the radius ratio, you determine how much potential difference exists based on the distribution of charge along the wire.

• The logarithmic term \(\ln\left( \frac{r_2}{r_1} \right)\) emphasizes how the potential difference changes with distance.
• As you move the probes further from each other, the potential difference reflects the strength and influence of the electric field.

Vacuum Permittivity

Vacuum permittivity, denoted as \(\varepsilon_0\), is a constant that appears in many electrostatic equations. Its value is approximately \(8.85 \times 10^{-12} \ \mathrm{C^2/N \, . \, m^2}\). This constant plays a crucial role in describing how electric fields interact in a vacuum.

In the exercise, \(\varepsilon_0\) is part of the formula used to calculate the potential difference. It influences how electric charge in free space affects other charges.

• \(\varepsilon_0\) helps define how strongly electric fields are set up between charges.
• It's an essential part of Coulomb's law, which describes the force between two point charges.
• Permittivity affects the strength and propagation of electric fields in various materials. For a vacuum, \(\varepsilon_0\) provides the baseline from which other materials' permittivities are compared.

Understanding vacuum permittivity helps make sense of how various dielectric materials would alter these fields when placed in them. This understanding can further inform us about materials' effectiveness in insulating electric charges.