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A very long conducting tube (hollow cylinder) has inner radius a and outer radius \(b\). It carries charge per unit length \(+a\), where \(a\) is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length \(+a\). (a) Calculate the electric field in terms of a and the distance r from the axis of the tube for (i) \(r < a; (ii) a < r < b; (iii) r > b\). Show your results in a graph of \(E\) as a function of \(r\). (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

Short Answer

Expert verified
For (i) \( E = \frac{a}{2\pi\varepsilon_0 r} \), (ii) \( E = 0 \), (iii) \( E = \frac{a}{\pi \varepsilon_0 r} \). Charges: inner \(-a\), outer \(+2a\).

Step by step solution

01

Understanding the Problem

We need to calculate the electric field for a conducting tube with specific charge distributions inside, between, and outside the tube. Then, analyze the charge distribution per unit length on the tube's surfaces.
02

Determine the Electric Field for r < a

Using Gauss's Law for a line charge located on the axis, the electric field only depends on the line of charge. For a cylindrical Gaussian surface of radius \( r < a \), \[E imes 2\pi r L = \frac{a L}{\varepsilon_0}\]where \( L \) is the length of the cylinder and \( E \) is the electric field, giving \[E = \frac{a}{2\pi \varepsilon_0 r}\].
03

Determine the Electric Field for a < r < b

Within this region, the conducting shell shields any external electric field, and by the properties of conductors, the electric field inside a charged conductor is zero. Thus, \[E = 0\].
04

Determine the Electric Field for r > b

Outside both the line charge and the charged tube, apply Gauss's Law again. The enclosed charge per unit length becomes \(2a\) (line charge and the total charge of the tube). For a cylindrical Gaussian surface of radius \( r > b \), \[E imes 2\pi r L = \frac{2a L}{\varepsilon_0}\]leading to\[E = \frac{2a}{2\pi \varepsilon_0 r} = \frac{a}{\pi \varepsilon_0 r}\].
05

Graph Electric Field as a Function of r

Plot the three sections: 1. For \( r < a \), \( E = \frac{a}{2\pi \varepsilon_0 r} \), decreases with \(r\).2. For \( a < r < b \), \( E = 0 \).3. For \( r > b \), \( E = \frac{a}{\pi \varepsilon_0 r} \), decreases with \(r\). The graph should reflect these changes in the electric field as a function of \( r \).
06

Charge per Unit Length on the Inner Surface of the Tube

The inner surface must neutralize the line charge to ensure zero electric field inside the conductor. Thus, to offset the \(+a\) on the axis, the inner surface holds charge per unit length \(-a\).
07

Charge per Unit Length on the Outer Surface of the Tube

The total charge per unit length given to the conducting cylindrical shell is \(+a\). The remaining charge, after the inner surface's neutrality, appears on the outer surface. Thus, the charge per unit length on the outer surface is \(+2a\).

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in electromagnetism, describing the force per unit charge experienced by a positive test charge at any point in space. It indicates the direction in which a positive charge would move if placed within this field.

Here's how the electric field varies in our problem involving a conducting cylinder with a line charge at its axis:
  • **Inside the cylinder (\(r < a\))**: The electric field is influenced only by the axial line charge, not by the conducting shell. Applying Gauss's Law, we find the electric field strength has an inverse relationship with the radius, expressed as \(E = \frac{a}{2\pi \varepsilon_0 r}\).
  • **Within the conductive shell (\(a < r < b\))**: The electric field is zero. This silence in the field occurs because the electric field inside a conductor in equilibrium state is always zero due to free charge movement that cancels it out.
  • **Outside the cylinder (\(r > b\))**: Here, both the line charge and the entire charge of the tube affect the field, giving \(E = \frac{a}{\pi \varepsilon_0 r}\). This field decreases with increasing distance.
Understanding these differences is crucial for calculating forces or potential energy within these spaces.
Conducting Cylinder
A conducting cylinder plays a special role in electromagnetism due to its ability to conduct electricity and distribute charge uniformly over its surface. In the context of our problem:

- **Uniqueness of Conductors**: In electrostatic equilibrium, the electric field inside a conductor is zero, a property that simplifies calculations and influences how charges distribute.

- **Charge Distribution**: Charge on a conductor redistributes itself across the surface to maintain zero electric field inside. For instance, a net charge on a conducting cylinder will spread across the surface, primarily residing on the outer surface.

In our exercise, these properties allowed us to determine that the electric field within the conductive material is zero, and the charge distribution adjusts to neutralize internal fields.
Charge Distribution
Charge distribution is a critical concept in electrostatics, determining how charges spread across conducting surfaces or within materials.

In our cylinder problem, we encounter different charge distribution scenarios:
  • **Line Charge**: Concentrates along the axis, contributing significantly to the electric field inside the cylinder.
  • **Inner Surface**: Must always balance any internal axial charges to maintain the conductor's equilibrium state. Here, it bears a charge per unit length of \(-a\), compensating for the positive axial line charge.
  • **Outer Surface**: Accumulates any remaining surplus or deficit charge, completing the overall charge distribution for the cylinder. In this case, it carries a charge per unit length of \(+2a\).
These distributions not only maintain the conductor's internal field at zero but also affect the external field.
Line Charge
A line charge is a model used to describe a uniformly distributed charge along a very long thin wire or straight line. This linear charge density helps simplify many physical problems.

- **Impact on Electric Field**: It creates a specific pattern of electric field lines, radiating symmetrically outward. This radial field simplifies application of Gauss’s law during calculations.

- **Calculation of Electric Field**: For the line charge in our problem, the field changes based on the radial distance from the line. Within the tube section (\(r < a\)), it contributes as the sole source of the electric field, calculated using Gauss’s Law. As we move outside the tube (\(r > b\)), it still affects the composite field along with charges from the conducting cylinder.

Understanding line charge properties is crucial in identifying how they influence fields in surrounding spaces, especially in conducting environments.

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Most popular questions from this chapter

The electric field at a distance of 0.145 m from the surface of a solid insulating sphere with radius 0.355 m is 1750 N/C. (a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? (b) Calculate the electric field inside the sphere at a distance of 0.200 m from the center.

In one experiment the electric field is measured for points at distances \(r\) from a uniform line of charge that has charge per unit length \(\lambda\) and length \(l\), where \(l \gg r\). In a second experiment the electric field is measured for points at distances \(r\) from the center of a uniformly charged insulating sphere that has volume charge density \(\rho\) and radius \(R =\) 8.00 mm, where \(r > R\). The results of the two measurements are listed in the table, but you aren't told which set of data applies to which experiment: For each set of data, draw two graphs: one for \(Er^2\) versus r and one for \(Er\) versus \(r\). (a) Use these graphs to determine which data set, A or B, is for the uniform line of charge and which set is for the uniformly charged sphere. Explain your reasoning. (b) Use the graphs in part (a) to calculate \(\lambda\) for the uniform line of charge and \(\rho\) for the uniformly charged sphere.

A very small object with mass 8.20 \(\times\) 10\(^{-9}\) kg and positive charge 6.50 \(\times\) 10\(^{-9}\) C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90 \(\times\) 10\(^{-8}\) C/m2. The object is initially 0.400 m from the sheet. What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.100 m?

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area \(\sigma =\) 5.00 \(\times\) 10\(^{-6}\) C/m\(^2\). (a) A small sphere of mass \(m =\) 8.00 \(\times\) 10\(^{-6}\) kg and charge \(q\) is placed 3.00 cm above the sheet of charge and then released from rest. (a) If the sphere is to remain motionless when it is released, what must be the value of \(q\)? (b) What is \(q\) if the sphere is released 1.50 cm above the sheet?

A point charge \(q_1 =\) 4.00 nC is located on the \(x\)-axis at \(x =\) 2.00 m, and a second point charge \(q_2 = -\)6.00 nC is on the \(y\)-axis at \(y =\) 1.00 m. What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m?

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