What is the Electric Field Inside and Outside a Charged Conducting Hollow Tube?

In summary, the problem involves a conducting tube with inner radius a and outer radius b, carrying a charge per unit length of +\alpha. Along the axis of the tube is a line of charge with the same charge per unit length. The electric field for r<a can be calculated using the formula E= (1/4*pi-epsilon_0)(q/R^2), where q is the total charge on the line of charge and R is the distance from the axis. For a<r<b, the electric field can be calculated in a similar manner. For r>b, the electric field falls off, but can still be calculated using the same formula. The charge per unit length on the inner surface of the tube is also +\
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jimbo71
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Homework Statement


A very long conducting tube (hollow cylinder) has inner radius a and outer radius b. It carries charge per unit length + \alpha , where \alpha 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 + \alpha .

Calculate the electric field in terms of \alpha and the distance r from the axis of the tube for r<a.
Express your answer in terms of the variables \alpha, r, and constant \epsilon_0.

Calculate the electric field in terms of \alpha and the distance r from the axis of the tube for a<r<b.
Express your answer in terms of the variables \alpha, r, and constant \epsilon_0.

Calculate the electric field in terms of \alpha and the distance r from the axis of the tube for r>b.
Express your answer in terms of the variables \alpha, r, and constant \epsilon_0.

What is the charge per unit length on the inner surface of the tube?

What is the charge per unit length on the outer surface of the tube?


Homework Equations


E=(1/4*pi-epsilon_0)(q/R^2)




The Attempt at a Solution


No idea where to start or how to set up problem
 
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To solve this problem, we can use the Gauss's Law which states that the electric field at a distance r from a line of charge is given by E = (1/4*pi*epsilon_0) * (q/l), where q is the total charge on the line and l is the length of the line. In this case, we have a conducting tube with charge per unit length +\alpha, so we can rewrite the equation as E = (1/4*pi*epsilon_0) * (+\alpha * l).

For r < a, the electric field is given by E = (1/4*pi*epsilon_0) * (+\alpha * l) = (1/4*pi*epsilon_0) * (+\alpha * 2*pi*r) = (+\alpha/2*epsilon_0) * r, where we have replaced the length of the line, l, with the circumference of the tube, 2*pi*r. This means that the electric field is directly proportional to the distance from the axis of the tube, r, and the charge per unit length, +\alpha.

For a < r < b, the electric field is given by E = (1/4*pi*epsilon_0) * (+\alpha * l) = (1/4*pi*epsilon_0) * (+\alpha * 2*pi*a) = (+\alpha/2*epsilon_0) * a, since the length of the line is still 2*pi*r, but now the charge is distributed over a smaller distance, a.

For r > b, the electric field is given by E = (1/4*pi*epsilon_0) * (+\alpha * l) = (1/4*pi*epsilon_0) * (+\alpha * 2*pi*b) = (+\alpha/2*epsilon_0) * b, since the length of the line remains the same, but now the charge is distributed over a larger distance, b.

To find the charge per unit length on the inner surface of the tube, we can use the fact that the electric field inside a conductor is zero. This means that the charge on the inner surface of the tube must be equal and opposite to the charge on the line of charge, +\alpha. Therefore, the charge per unit length on the inner surface is -\alpha.

Similarly, for the outer surface of the
 

Related to What is the Electric Field Inside and Outside a Charged Conducting Hollow Tube?

1. What is a conducting hollow tube?

A conducting hollow tube is a tube made of a conductive material, such as metal, that allows the flow of electricity or heat through its interior. It can also refer to a tube used in scientific experiments to conduct electricity or heat.

2. What are the properties of a conducting hollow tube?

A conducting hollow tube has the ability to conduct electricity or heat, as well as the ability to maintain its shape and structure while allowing the flow of these elements. It is also typically resistant to corrosion and can have high strength and durability.

3. How is a conducting hollow tube used in scientific experiments?

A conducting hollow tube is commonly used in experiments to study the flow of electricity or heat through different materials. It can also be used as a part of a circuit or to create a specific environment for conducting experiments.

4. What are the advantages of using a conducting hollow tube?

One advantage of using a conducting hollow tube is its ability to conduct electricity or heat efficiently. It can also be easily shaped and manipulated for specific experimental needs. Additionally, it is often more durable and long-lasting compared to other materials, making it suitable for repeated use.

5. How do you choose the right conducting hollow tube for an experiment?

The right conducting hollow tube for an experiment will depend on the specific properties and needs of the experiment. Factors to consider include the material, size, and shape of the tube, as well as the type and amount of electricity or heat that needs to be conducted. It is important to carefully evaluate these factors and choose a tube that will best suit the experiment's objectives.

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