Problem 32 - University Physics with Modern Physics

Question

A very long insulating cylinder of charge of radius 2.50 \(\mathrm{cm}\) carries a uniform linear density of 15.0 \(\mathrm{nC} / \mathrm{m} .\) 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 \(\mathrm{V} ?\)

Step-by-Step Solution

Answer

The probe should be placed 0.39 m from the surface.

Step 1: Understanding the Problem

The problem involves finding the distance from the surface of a charged cylindrical insulator where the electric potential difference is 175 V. We have a cylinder with a radius of 2.50 cm and a linear charge density \( \lambda = 15.0 \text{ nC/m} \) (which is \( 15.0 \times 10^{-9} \text{ C/m} \)). We are asked to determine the distance from the cylinder's surface to a point where the electric potential is 175 V greater than at the surface.

Step 2: Determine Electric Field

For a long charged cylinder, the electric field at a distance \( r \) from its axis is given by Gauss's Law: \[ E(r) = \frac{\lambda}{2 \pi \varepsilon_0 r} \] where \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{(N m)}^2 \).

Step 3: Calculate Potential Difference

The potential difference \( \Delta V \) between two points is the negative integral of the electric field \( E \) from \( r_1 \) to \( r_2 \): \[ \Delta V = - \int_{r_1}^{r_2} E(r) \, dr = - \int_{r_1}^{r_2} \frac{\lambda}{2 \pi \varepsilon_0 r} \, dr = \frac{\lambda}{2 \pi \varepsilon_0} \ln\left( \frac{r_2}{r_1} \right) \] where \( r_1 \) is the radius of the cylinder (2.50 cm). \( \Delta V = 175 \text{ V} \).

Step 4: Solve for \( r_2 \)

Rearrange the equation to solve for \( r_2 \): \[ 175 = \frac{15.0 \times 10^{-9}}{2 \pi \times 8.85 \times 10^{-12}} \ln\left( \frac{r_2}{0.025} \right) \] Simplify and solve for \( r_2 \): \[ r_2 \approx 0.415 \text{ m} \]

Step 5: Calculate Distance from Surface

The problem asks for the distance from the surface of the cylinder. Since \( r_1 = 2.50 \text{ cm} = 0.025 \text{ m} \) and we found \( r_2 = 0.415 \text{ m} \), the distance from the surface is \( r_2 - r_1 = 0.415 \text{ m} - 0.025 \text{ m} = 0.39 \text{ m} \).

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle relating electric fields to charge distributions. Imagine you have a long charged cylinder, like the one in our exercise.
To determine the electric field around it, Gauss's Law is your go-to resource. Gauss's Law states that the total electric flux through a closed surface is equal to the charge enclosed divided by the electric constant, \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]When applied to symmetric shapes like spheres or cylinders, solving these equations becomes much easier. In our case, the electric field at any point outside the cylinder depends only on the radius from the cylinder's axis, not on the height or the angular position around the axis. This cylindrical symmetry simplifies calculations significantly. Using Gauss's Law, an electric field surrounding a uniformly charged cylinder can be expressed as:\[ E(r) = \frac{\lambda}{2 \pi \varepsilon_0 r} \]Here, \( \lambda \) is the linear charge density. This formula helps us find the electric field strength at different radial distances from the cylinder's axis.

Linear Charge Density

Linear charge density is essentially the amount of electric charge per unit length along a line. It's represented by the Greek letter lambda (\( \lambda \)).
Think about it this way: if you have a long cylinder with charge spread evenly along its length, then the charge at any point along the length is determined by \( \lambda \).In our exercise, the linear charge density of the cylinder is given as 15.0 nC/m. This means that every meter of the cylinder carries 15.0 nanocoulombs of charge, where 1 nanocoulomb (nC) is equal to \( 10^{-9} \text{ C} \). Knowing \( \lambda \) is crucial because it allows us to calculate the electric field using Gauss's Law, and further compute the electric potential difference between points around the cylinder. Linear charge density is particularly useful when dealing with problems involving long cylindrical shapes since such distributions are typically uniform, simplifying calculations. Always keep in mind that understanding \( \lambda \) is key to unlocking the behavior of electric fields and potentials around charged wires or rods.

Electric Field of a Cylinder

The electric field of a cylinder is one of those concepts that align beautifully with symmetry principles like Gauss's Law. For a charged cylinder, the electric field outside the cylinder decreases with distance from the center.
When dealing with an infinitely long charged cylinder, it's usually symmetric around its length. We determine the electric field by focusing on a Gaussian surface—a hypothetical closed surface—aligned around the cylinder's length. This doesn't rely on height since the field is evenly distributed circumferentially. The expression for the electric field around our cylindrical charge is:\[ E(r) = \frac{\lambda}{2 \pi \varepsilon_0 r} \]In essence, the field strength decreases inversely with the distance \( r \). For our problem, using this field expression allows us to manage potential differences, which link directly to how the electric field strength varies with distance. An important aspect is that this is only valid outside the cylinder's radius—inside, different rules apply. Understanding these field expressions gives us insight into how electricity behaves around linear charged objects, and helps solve problems dealing with potential differences, as seen in our exercise.