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A hollow, thin-walled insulating cylinder of radius \(R\) and length \(L\) (like the cardboard tube in a roll of toilet paper) has charge \(Q\) uniformly distributed over its surface. (a) Calculate the electric potential at all points along the axis of the tube. Take the origin to be at the center of the tube, and take the potential to be zero at infinity. (b) Show that if \(L \ll R\), the result of part (a) reduces to the potential on the axis of a ring of charge of radius \(R\). (See Example 23.11 in Section 23.3.) (c) Use the result of part (a) to find the electric field at all points along the axis of the tube.

Short Answer

Expert verified
Potential on axis: \( V = \frac{kQ}{L} \ln\left(\frac{z+\sqrt{R^2+z^2}+L/2}{z+\sqrt{R^2+z^2}-L/2}\right) \); if \(L \ll R\), it simplifies to a ring's potential. Electric field: \( E = -\frac{dV}{dz} \).

Step by step solution

01

Identify the setup of the problem

We have a hollow, thin-walled insulating cylinder with charge \(Q\) uniformly distributed over its surface. The cylinder has radius \(R\) and length \(L\). We need to find the electric potential along the axis of the tube, assuming potential is zero at infinity.
02

Derive the expression for the potential along the axis (Part a)

Considering a small charge element \(dq\) on the surface of the cylinder, at a distance \(z\) from the point along the axis where the potential is being calculated. The linear charge density \(\lambda = \frac{Q}{L}\). For a point at distance \(z\) along the axis, the potential \(dV\) due to \(dq\) is \(dV = \frac{k \cdot dq}{\sqrt{R^2 + z^2}}\). Integrate over the length of the cylinder (from \(-\frac{L}{2}\) to \(\frac{L}{2}\)): \[V = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{k \lambda R \ d\theta}{\sqrt{R^2 + z^2}} = \frac{kQ}{L}\int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{1}{\sqrt{R^2 + z^2}} \, dz.\] Solving this integral, we find: \[ V = \frac{kQ}{L} \ln\left(\frac{z + \sqrt{R^2 + z^2} + \frac{L}{2}}{z + \sqrt{R^2 + z^2} - \frac{L}{2}}\right). \]
03

Simplify for \(L \ll R\) and compare with a ring of charge (Part b)

If \(L \ll R\), the cylinder approximates a ring of charge, and the length \(L\) becomes negligible compared to \(R\). The expression simplifies to the potential on axis of a ring of charge:\[ V = \frac{kQ}{\sqrt{R^2 + z^2}}. \] This matches the expected formula for a ring of charge.
04

Derive the electric field along the axis (Part c)

The electric field is the negative gradient of potential, so we take the derivative of the potential \(V\) with respect to \(z\). Using the earlier derived potential: \[E = -\frac{dV}{dz} = -\frac{kQ}{L}\left(\frac{1}{\sqrt{R^2 + (z+\frac{L}{2})^2}} - \frac{1}{\sqrt{R^2 + (z-\frac{L}{2})^2}}\right).\] This gives the electric field along the axis of the cylinder.

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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 physics, representing the force that a charged object exerts on any other charge in its vicinity. Think of it like an invisible web of influence surrounding a charge.
If you place another charge within this web, that second charge will experience a force without actually touching the first charge. This field is represented by vectors that illustrate the direction and magnitude of the force felt by a positive test charge.

Mathematically, the electric field \(E\) can be derived from the electric potential \(V\). It is related to the potential by the gradient where \[E = -\frac{dV}{dz}\text{ (for one-dimensional motion)}.\] It signifies that the electric field direction is steepest in the direction where potential decreases fastest. When analyzing fields, especially for geometrically symmetric objects like cylinders, the field typically points perpendicular to surfaces and is strongest where lines are closest together. This way, calculating and understanding the potential helps find the electric field in complex arrangements, like those of cylinders of charge.
Cylinder of Charge
A cylinder of charge is a common model in electrostatics. Imagine a long hollow tube that has charge spread evenly over its surface. This setup helps when studying fields and potentials due to symmetrical charge distributions.
With a cylinder, variables like radius \(R\), length \(L\), and total charge \(Q\) matter tremendously. These parameters define how charges are spread and influence points within or outside the cylinder.

Working with cylinders simplifies complex geometrical models. By breaking the problem into smaller components, like infinitesimally small charge elements, students can use integration to determine forces, fields, and potentials under various conditions.
Uniform Charge Distribution
Uniform charge distribution means that each piece of the charged object carries the same amount of charge per unit area or length. It's like spreading peanut butter smoothly over a slice of bread - every part of the bread gets an equal amount of peanut butter. In our hollow cylinder, this uniform spread means that each small segment of the surface carries \(\lambda = \frac{Q}{L}\), where \(\lambda\) is the linear charge density.

This uniformity ensures that calculations for potential and electric fields can take advantage of symmetry. This leads to simpler integrals and more straightforward math. Without it, finding fields or potentials would be more complex and less likely to have elegant solutions. Hence, understanding uniform distribution is key to efficiently solving electrostatic problems.
Integral Calculus
Integral calculus is crucial when dealing with continuous charge distributions, like our thin-walled cylinder. It's a part of calculus focused on accumulation, telling us the total effect of adding up many small pieces, which is essential for electromagnetism.
By integrating over a charge distribution, one can find the total potential or field at a point. For a cylinder of charge, this involves integration over each element of the cylinder, considering how each affects a point on its axis.

The basic principle: break down the charge into infinitesimal parts \(dq\), compute their individual potentials, and then sum these. For instance, the integration for potential involves the equation:
\[ V = \int \frac{k \cdot dq}{\sqrt{R^2 + z^2}} \] which tells us how the entire charged cylinder influences a point along its axis.
Ring of Charge Approximation
When a cylinder's length \(L\) is much smaller than its radius \(R\) (\(L \ll R\),), it can be approximated as a ring of charge. This is because the length becomes negligibly small compared to how far the charge reaches around the cylindrical curve.
Approximating the cylinder as a ring simplifies calculations since evaluating the potential or field from a ring is significantly easier due to its inherent symmetry.
  • The potential for a ring is derived more simply as:
    \[ V = \frac{kQ}{\sqrt{R^2 + z^2}} \] which illustrates how potent the symmetry of a ring can be in simplifying electrostatic calculations.
This approach is a powerful tool to make complex problems more tractable by trading off some precision for conceptual simplicity, making it a cornerstone in physics problem-solving.

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Most popular questions from this chapter

An alpha particle with kinetic energy 9.50 MeV (when far away) collides head- on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

When radium-226 decays radioactively, it emits an alpha particle (the nucleus of helium), and the end product is radon-222. We can model this decay by thinking of the radium-226 as consisting of an alpha particle emitted from the surface of the spherically symmetric radon-222 nucleus, and we can treat the alpha particle as a point charge. The energy of the alpha particle has been measured in the laboratory and has been found to be 4.79 MeV when the alpha particle is essentially infinitely far from the nucleus. Since radon is much heavier than the alpha particle, we can assume that there is no appreciable recoil of the radon after the decay. The radon nucleus contains 86 protons, while the alpha particle has 2 protons and the radium nucleus has 88 protons. (a) What was the electric potential energy of the alpha\(-\)radon combination just before the decay, in MeV and in joules? (b) Use your result from part (a) to calculate the radius of the radon nucleus.

The maximum voltage at the center of a typical tandem electrostatic accelerator is 6.0 MV. If the distance from one end of the acceleration tube to the midpoint is 12 m, what is the magnitude of the average electric field in the tube under these conditions? (a) 41,000 V/m; (b) 250,000 V/m; (c) 500,000 V/m; (d) 6,000,000 V/m.

A point charge \(q_1 = +\)2.40 \(\mu\)C is held stationary at the origin. A second point charge \(q_2 = -\)4.30 \(\mu\)C moves from the point \(x =\) 0.150 m, \(y =\) 0 to the point \(x =\) 0.250 m, \(y =\) 0.250 m. How much work is done by the electric force on \(q_2\)?

The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but opposite charges. Typical dimensions are about 3.0 cm on a side, with a separation of about 5.0 mm. The potential difference between the plates is 25.0 V. The plates are close enough that we can ignore fringing at the ends. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

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