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Zero field inside a cylindrical shell * Consider a distribution of charge in the form of a hollow circular cylinder, like a long charged pipe. In the spirit of Problem 1.17, show that the electric field inside the pipe is zero.

Short Answer

Expert verified
Using Gauss's Law, it's shown that the electric field inside a uniformly charged cylindrical shell is zero due to symmetry.

Step by step solution

01

Understand Gaussian cylinder

We'll take a Gaussian cylinder inside the hollow charged cylinder. The Gaussian cylinder is a hypothetical cylinder that is used here as part of the application of Gauss’s law. Note that this isn't a real object; it’s a surface we mentally impose on the real objects in the problem to help us apply Gauss’s law.
02

Apply Gauss's law

According to Gauss's Law, the electric field \(E\) times the area \(A\) equals the total electric charge \(Q\) inside divided by the permittivity of free space \(\epsilon_{0}\). Mathematically, it's expressed as \(E A = Q/\epsilon_{0}\). The total charge inside the Gaussian cylinder is zero because it’s within the hollow part of the charged cylindrical shell. So on the right side of the equation, we have zero.
03

Resolve Gauss's equation

Substituting \(Q = 0\) in Gauss's Law equation, we have \(E A = 0\). Therefore, this means that the electric field \(E = 0\) inside the cylindrical shell, proving the claim.

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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 vector quantity that represents the force per unit charge experienced by a small positive test charge placed in the vicinity of other charges. It's a fundamental concept in electromagnetism and plays a crucial role in understanding how charges interact with one another. The direction of the electric field at a point in space is the direction of the force that would act on a positive test charge placed at that point. For any given charge distribution, the electric field can be calculated at any point in space.

Using the concept of electric field, one can predict behaviours such as attraction or repulsion between charged objects and identify the influence of charges on nearby neutral objects. It's essential in solving problems related to electrical forces and is a cornerstone concept when using Gauss's Law.
Gaussian Surface
A Gaussian surface is an imaginary closed surface used in Gauss's Law applications, a fundamental principle in electrostatics. It’s chosen strategically to exploit symmetry, making the calculations more manageable. The magic of a Gaussian surface is that it lets you relate the electric field at various points on the surface to the total charge enclosed by that surface.

In the context of a charged cylindrical shell, a Gaussian surface might be a cylindrical surface co-axial with the shell that helps us analyze the electric field within, on, and outside the shell. The choice of this imaginary surface is not unique, but its orientation and shape are selected based on the symmetry of the charge distribution for simplifying the calculations.
Charged Cylindrical Shell
A charged cylindrical shell can be visualized as a hollow pipe that carries a uniform static charge spread along its surfaces but with no charge in its interior. When an exercise prompts us to prove zero electric field within such a shell, the cylindrical symmetry comes into play. This symmetry implies that the electric field at any point within the hollow part must be radial and at a constant magnitude for each specific distance from the axis—leading us to consider a Gaussian surface that mirrors this symmetry.

The cylindrical shell's inner surface bears importance as well since it outlines the region where the electric field is being evaluated. Such problems typically assume an infinite shell length to minimize edge effects and make the symmetry perfect, but in practical applications, this idealization is still useful for long, thin shells.
Electric Charge
Electric charge is a fundamental property of particles that causes them to experience a force within an electric field. Charges are present in two types, positive and negative, and they govern the electrical properties of matter. The behaviour of electric charge is encapsulated by Coulomb's law, which describes how charged particles interact, and by Gauss's Law, which relates the electric field to charge distribution.

In our exercise concerning the charged cylindrical shell, we focus on the role of charge inside the Gaussian surface. The electric field inside the shell is zero when there's no net charge within the Gaussian surface—even though charges may be present elsewhere, like on the shell's surfaces.
Permittivity of Free Space
Permittivity of free space, denoted as \(\text{\(\epsilon_{0}\)}}\), is a physical constant that characterizes the ability of the vacuum to permit electric field lines. This constant appears in several fundamental equations in electromagnetism, like Coulomb’s law and Gauss's Law. It quantifies how much electric field is 'allowed' to emanate from a given electric charge in free space.

In our exercise with the cylindrical shell, \(\epsilon_{0}\) appears in the denominator of Gauss's Law. When there is no charge within the Gaussian surface, as is the case with our hollow cylindrical shell, the product of the electric field and area is set to zero, which relates directly to the absence of net charge within that surface, without directly involving \(\epsilon_{0}\) since it gets multiplied by zero.

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Most popular questions from this chapter

Decreasing energy? A hollow spherical shell with radius \(R\) has charge \(Q\) uniformly distributed over it. The task of Problem \(1.32\) is to show that the energy stored in this system is \(Q^{2} / 8 \pi \epsilon_{0} R\). (You can derive this here if you want, or you can just accept it for the purposes of this problem.) Now imagine taking all of the charge and concentrating it in two point charges \(Q / 2\) located at diametrically opposite positions on the shell. The energy of this new system is \((Q / 2)^{2} / 4 \pi \epsilon_{0}(2 R)=\) \(Q^{2} / 32 \pi \epsilon_{0} R\), which is less than the energy of the uniform spherical shell. Does this make sense? If not, where is the error in this reasoning?

Potential energy of a cylinder * * Problem \(1.24\) gives one way of calculating the energy per unit length stored in a solid cylinder with radius \(a\) and uniform volume charge density \(\rho .\) Calculate the energy here by using Eq. (1.53) to find the total energy per unit length stored in the electric field. Don't forget to include the field inside the cylinder. You will find that the energy is infinite, so instead calculate the energy relative to the configuration where all the charge is initially distributed uniformly over a hollow cylinder with large radius \(R\). (The field outside radius \(R\) is the same in both configurations, so it can be ignored when calculating the relative energy.) In terms of the total charge \(\lambda\) per unit length in the final cylinder, show that the energy per unit length can be written as \(\left(\lambda^{2} / 4 \pi \epsilon_{0}\right)(1 / 4+\ln (R / a))\).

Field in the end face Consider a half-infinite hollow cylindrical shell (that is, one that extends to infinity in one direction) with uniform surface charge density. Show that at all points in the circular end face, the electric field is parallel to the cylinder's axis. Hint: Use superposition, along with what you know about the field from an infinite (in both directions) hollow cylinder.

Two equal fields * The result of Exercise \(1.78\) is that the electric field at the center of a small hole in a spherical shell equals \(\sigma / 2 \epsilon_{0}\). This happens to be the same as the field due to an infinite flat sheet with the same density \(\sigma\). That is, at the center of the hole at the top of the spherical shell in Fig. 1.39, the field from the shell equals the field from the infinite horizontal sheet shown. (This sheet could actually be located at any height.) Demonstrate this equality by explaining why the rings on the shell and sheet that are associated with the angle \(\theta\) and angular width \(d \theta\) yield the same field at the top of the shell.

Field from two charges ** A charge \(2 q\) is at the origin, and a charge \(-q\) is at \(x=a\) on the \(x\) axis. (a) Find the point on the \(x\) axis where the electric field is zero. (b) Consider the vertical line passing through the charge \(-q\), that is, the line given by \(x=a\). Locate, at least approximately, a point on this line where the electric field is parallel to the \(x\) axis.

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