Question

A very long insulating cylinder of charge of radius 2.50 cm carries a uniform linear density of 15.0 nC\(/\)m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175 V?

Short answer

The probe must be 0.12 m from the surface.

Step-by-step solution

Understand the Problem

We need to find the distance from the surface of a charged cylinder where the potential difference is 175 V. The charge distribution is uniform, and the problem involves electric fields and potentials outside a charged cylinder.

Define the Electric Field Outside the Cylinder

For a very long charged cylinder with linear charge density \( \lambda \), the electric field at a distance \( r \) from the axis is given by Gauss's Law: \[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \]where \( \varepsilon_0 \) is the permittivity of free space.

Calculate the Potential Difference Expression

The potential difference \( V \) between two points at distances \( r_1 \) and \( r_2 \) from the axis is obtained by integrating the electric field:\[ V = -\int_{r_1}^{r_2} E \, dr = -\int_{r_1}^{r_2} \frac{\lambda}{2\pi\varepsilon_0 r} \, dr = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_2}{r_1}\right) \]

Enter Known Values

From the cylinder's surface, \( r_1 = 2.50 \times 10^{-2} \) m, and the linear charge density \( \lambda = 15.0 \times 10^{-9} \) C/m. The potential difference \( V = 175 \) V. Plug these into the formula:\[ 175 = \frac{15.0 \times 10^{-9}}{2\pi \times 8.85 \times 10^{-12}} \ln\left(\frac{r_2}{2.50 \times 10^{-2}}\right) \]

Solve for \( r_2 \)

Rearrange the formula to solve for \( r_2 \): \[ \ln\left(\frac{r_2}{2.50 \times 10^{-2}}\right) = \frac{175 \times 2\pi \times 8.85 \times 10^{-12}}{15.0 \times 10^{-9}} \approx 20.64 \]Thus,\[ \frac{r_2}{2.50 \times 10^{-2}} = e^{20.64} \]\[ r_2 = 2.50 \times 10^{-2} \times e^{20.64} \]

Calculate \( r_2 \, and \, Find \, the \, Distance \, from \, Surface

Calculate \( r_2 \), then find the distance from the surface:\[ r_2 \approx 2.50 \times 10^{-2} \times e^{20.64} \approx 0.145 \text{ m} \]Distance from surface = \( r_2 - 0.025 \approx 0.12 \text{ m} \).

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism. It states that the electric flux through a closed surface is proportional to the enclosed electric charge. This law helps us understand the distribution of electric fields around charged objects.
The law is mathematically expressed as:

• \( \Phi = \frac{Q}{\varepsilon_0} \)

where \( \Phi \) represents the electric flux, \( Q \) is the total charge enclosed, and \( \varepsilon_0 \) is the permittivity of free space. For a long, charged cylinder, Gauss's Law helps us derive the electric field outside the cylinder, which is essential in determining potential differences. The symmetry of the cylinder allows us to apply Gauss's Law effectively by using a cylindrical Gaussian surface, leading to a simple expression for the electric field. Understanding this concept is key to solving problems involving cylindrical charge distributions.

Electric Field

The electric field is a vector field that surrounds charged objects. It describes the force a charged particle feels within the field's presence. For a charged cylinder, the electric field outside can be derived using Gauss's Law. It is particularly important to understand that outside a long cylinder of uniform charge density, the electric field diminishes with the increase in distance from its surface.
The formula derived from Gauss's Law for the electric field around a charged cylinder is:

• \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \)

where:

• \( E \) is the electric field,
• \( \lambda \) is the linear charge density,
• \( r \) is the radius or distance from the axis of the cylinder.

This relationship indicates that the electric field is inversely proportional to the distance \( r \). As the distance increases, the electric field decreases. Understanding this inverse relationship is crucial when analyzing electric potentials and fields around cylindrical shapes.

Linear Charge Density

Linear charge density, denoted as \( \lambda \), is a measure of electric charge per unit length along a line, and it's particularly useful when dealing with objects like wires or cylinders. It is calculated by dividing the total charge by the length of the object.
In this context, the linear charge density is critical for evaluating the electric field and potential difference around the charged cylinder.
Key points to remember about linear charge density:

• \( \lambda = \frac{Q}{L} \)
• \( Q \) is the total charge,
• \( L \) is the length of the charged cylinder.
• Unit is typically coulombs per meter (C/m).

In our example, a linear charge density of 15.0 nC/m means that every meter length of the cylinder holds 15.0 nC of charge. This simplifies our calculations of electric fields using Gauss's Law and is essential when determining how potential differences arise within the field generated by the cylinder.

Integration of Electric Fields

Integration of electric fields is a valuable technique in electric potential calculations. It involves integrating the electric field over a specified path to determine the potential difference between two points. For potential differences related to a cylindrical charge, this method is indispensable.
The integral of the electric field is:

• \( V = -\int_{r_1}^{r_2} E \, dr \)

where \( V \) is the potential difference, and \( r_1 \) and \( r_2 \) are the initial and final radii from the cylinder's axis. By solving this integral, we obtain:

• \( V = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_2}{r_1}\right) \)

This expression tells us how the potential difference is related to the charge density and distance. It's crucial for answering practical questions like how far an electric potential will extend from a surface, as seen in the given exercise about placing voltmeter probes.