91Ó°ÊÓ

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? Given: \(\ln \left(\frac{7}{5}\right)=0.337\)

Step by step solution

01

Understand the Electric Potential Due to a Line Charge

The electric potential difference between two points a distance \( r_1 \) and \( r_2 \) from an infinite line charge with linear charge density \( \lambda \) is given by: \[ \Delta V = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_2}{r_1}\right) \]\where \( \varepsilon_0 \) is the vacuum permittivity, approximately \( 8.85 \times 10^{-12} \mathrm{~C^2/Nm^2} \).

02

Calculate \(\lambda\) for Given Potential Difference

Given one probe is placed at \( r_1 = 2.50 \ \mathrm{cm} = 0.025 \ \mathrm{m} \) and the other at \( r_2 = 3.50 \ \mathrm{cm} = 0.035 \ \mathrm{m} \), with a potential difference of \( 575 \, \mathrm{V} \):Set up the equation: \[ 575 = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{0.035}{0.025}\right) \]Solve for \( \lambda \):\[ \lambda = \frac{575 \times 2\pi\varepsilon_0}{\ln(\frac{7}{5})} \approx \frac{575 \times 2\pi \times 8.85 \times 10^{-12}}{0.337} \approx 1.91 \times 10^{-10} \, \mathrm{C/m} \]

03

Determine Potential for a New Probe Placement

When the probes are placed at \( 3.50 \, \mathrm{cm} = 0.035 \, \mathrm{m} \) and \( 4.50 \, \mathrm{cm} = 0.045 \, \mathrm{m} \):Calculate \( \Delta V \): \[ \Delta V = \frac{1.91 \times 10^{-10}}{2\pi \times 8.85 \times 10^{-12}} \ln\left(\frac{0.045}{0.035}\right) \]Using the approximation \( \ln\left(\frac{9}{7}\right) \approx 0.240 \), find:\[ \Delta V \approx 575 \frac{0.240}{0.337 \ln\left(\frac{9}{7}\right)} = 409 \, \mathrm{V}\]Thus, the voltmeter will not read 575 V; it will show less.

04

Calculate Potential with Probes at Equal Distance

When both probes are at \( 3.50 \, \mathrm{cm} = 0.035 \, \mathrm{m} \), they are equidistant, so theoretically, there should be no potential difference detected by the voltmeter since both points are at the same potential: \[ \Delta V = 0 \, \mathrm{V} \] Hence, the voltmeter should read 0 V because the same distance from the wire implies no potential difference.

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Key Concepts

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

Electric potential difference

Electric potential difference is a concept that measures how much energy is needed to move a charge between two points in an electric field. In simpler terms, it's like the difference in effort needed to carry a heavy object uphill to two different heights; you expend more effort to get to a higher point. Electric potential difference is measured in volts (V). For a line charge, this difference depends on the distances from the charge. Using a voltmeter, we can determine this difference between two points along the wire.

• The readings depend on the distances from the line charge.
• Potential difference is higher if the second point is further from the line charge compared to the first.

In our exercise with a specific line charge, the voltmeter reads 575 V when placed at certain points from the wire. It would read differently if the points were changed due to the variation in potential difference.

Linear charge density

Linear charge density (\(\lambda\)) is a measure of how much electrical charge is distributed along a line, like a long wire. Imagine spreading sugar evenly along a narrow path; linear charge density is similar but with electrical charge. It is given in terms of charge per unit length, measured in coulombs/meter (C/m).

• In the given problem, a very long wire has a uniform linear charge density \(\lambda\).
• We calculate \(\lambda\) using the potential difference formula and given distances.

By measuring the potential difference with probes in different positions relative to the wire, you can solve for \(\lambda\), revealing how much charge is actually spread over the length of the wire.

Vacuum permittivity

Vacuum permittivity, denoted as \(\varepsilon_0\), is a fundamental physical constant that describes how electric fields interact in a vacuum. Think of it as a measure of the "ease" with which an electric field can establish itself in free space. It plays a crucial role in calculations involving electric fields and forces, and is approximately \(8.85 \times 10^{-12} \ \mathrm{C^2/Nm^2}\).

• It helps in determining the electric field caused by a charged object in a vacuum.
• In our context, it's used in the formula for electric potential due to a linear charge.

In the problem, vacuum permittivity is a part of the formula that calculates the potential difference around the wire, helping us solve for the linear charge density \(\lambda\). Without \(\varepsilon_0\), this calculation wouldn't be possible.

Logarithmic potential calculation

Logarithmic potential calculations involve using logarithms to solve problems related to potential differences in physics. It applies when dealing with infinite or very long charged objects like wires. The potential difference formula for a line charge makes use of the natural logarithm (\(\ln\)) to determine how the potential changes with distance from the charge.

• Logarithmic calculations help simplify complex relationships between distance and potential.
• In the given exercise, the potential difference was calculated using the natural logarithm of the distance ratio \(\ln\left(\frac{r_2}{r_1}\right)\).

This calculation is crucial as it incorporates how the potential difference diminishes with increasing distance from the line charge. Understanding this concept allows for determining potential differences in systems with more complex geometries over longer distances.