Chapter 4: Problem 10
Assuming that a beam of electrons can be treated as a charge distributed uniformly over a cylindrical region of radius \(a\) and of infinite length, find the Efield at a distance \(r\) from the axis for \(r>a\) and \(r
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01
Applying Gauss's Law to find the electric field
We will start by applying Gauss's law for the two cases mentioned above (\(r > a\) and \(r < a\)). Gauss's law states that the total electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (\(\varepsilon_{0}\)):
\(\oint \vec{E} \cdot d\vec{A} = \dfrac{Q_{enclosed}}{\varepsilon_{0}}\)
For a cylindrical Gaussian surface of radius \(r\) and height \(h\), we can simplify Gauss's law to:
\(E(2\pi r h) = \dfrac{\lambda h}{\varepsilon_{0}}\), where \(\lambda\) is the linear charge density.
Now, we have two cases for our cylindrical charge distribution:
For \(ra\): The electric field \(E_{2}\) for this case can be simplified to:
\(E_{2}(2\pi r h) = \dfrac{\lambda_{2} h}{\varepsilon_{0}}\)
02
Solving for the electric fields E1 and E2
We will now solve for the electric fields for both cases:
For \(ra\): Since the Gaussian surface encloses the entire cylinder of charge with radius \(a\) for this case, \(\lambda_{2}=\lambda\). Substituting this into our simplified Gauss's law equation:
\(E_{2}(2\pi r h) = \dfrac{\lambda h}{\varepsilon_{0}}\)
Simplifying for \(E_{2}\), we get:
\(E_{2} = \dfrac{\lambda}{2\pi r\varepsilon_{0}}\) for \(r>a\)
03
Finding the force on an electron
Now that we have the expressions for the electric field at a distance \(r\) from the axis, we can find the force on an electron at that distance. The force is given by:
\(F = qE\), where \(q\) is the charge of the electron.
Since there are \(n\) electrons per unit volume, the force on a single electron in the beam can be written as:
\(F_{e} = neE\)
We'll use the expressions for \(E_1\) and \(E_2\) that we derived above to calculate the force on the electron at a distance \(r\) from the axis:
For \(ra\): \(F_{e2} = neE_{2} = \dfrac{ne\lambda}{2\pi r\varepsilon_{0}}\)
Now, we know that \(e\) is the electronic charge and \(\lambda = n e a^2\), substituting this value of \(\lambda\) in both equations:
For \(ra\): As \(r>a\), the electron is outside the cylindrical region. Hence there is no electric field and the force on the electron is zero, so \(F_{e2}=0\).
Thus, the force on an electron at a distance \(r\) from the axis is \(F_{e} = \dfrac{ne^2 r}{2\varepsilon_{0}}\) (only if \(ra\)) radially outwards.
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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 representing the force that would be exerted on a unit charge placed at a point in space. It is a vector field that is characterized by both magnitude and direction.
When we discuss the electric field due to various charge distributions, we often invoke Gauss's Law, which provides a means to calculate the electric field by relating it to the charge within a certain region. For the case of a cylindrical distribution of charge, the symmetry of the problem simplifies the electric field's direction to be purely radial, that is, away from or towards the axis of the cylinder, depending on the sign of the charge. Thus, using Gauss's Law in such symmetrical situations, we can calculate the electric field without integrating by considering a Gaussian surface that suits the symmetry of the charge distribution.
This is particularly useful when considering the force experienced by particles within these fields, such as electrons in a beam. The force on a charged particle is simply the product of the electric field and the charge of the particle, as shown in the exercise. It's this interplay between electric fields and charges that underlies all electrostatic phenomena.
When we discuss the electric field due to various charge distributions, we often invoke Gauss's Law, which provides a means to calculate the electric field by relating it to the charge within a certain region. For the case of a cylindrical distribution of charge, the symmetry of the problem simplifies the electric field's direction to be purely radial, that is, away from or towards the axis of the cylinder, depending on the sign of the charge. Thus, using Gauss's Law in such symmetrical situations, we can calculate the electric field without integrating by considering a Gaussian surface that suits the symmetry of the charge distribution.
This is particularly useful when considering the force experienced by particles within these fields, such as electrons in a beam. The force on a charged particle is simply the product of the electric field and the charge of the particle, as shown in the exercise. It's this interplay between electric fields and charges that underlies all electrostatic phenomena.
Cylindrical Charge Distribution
Cylindrical charge distribution refers to a scenario where charge is spread out uniformly along a cylinder's surface or throughout its volume. Due to its symmetry, we can use Gauss's Law quite effectively here. The electric field is easier to calculate at points along or outside the axis of the cylinder, especially when the cylinder is assumed to be infinitely long. In real-world scenarios, this kind of distribution can be seen in certain types of capacitors or in beams of charged particles, such as the electron beam in the exercise.
For an infinite cylindrical charge distribution, the electric field outside the cylinder (r > a) depends on the linear charge density and inversely on the distance from the center of the cylinder. Inside (r < a), the calculation takes into account how much of the charge distribution is enclosed by the Gaussian surface. This leads to different expressions for the electric field depending on the position with respect to the cylinder's radius, and it highlights how geometry plays a crucial role in determining electrostatic effects.
For an infinite cylindrical charge distribution, the electric field outside the cylinder (r > a) depends on the linear charge density and inversely on the distance from the center of the cylinder. Inside (r < a), the calculation takes into account how much of the charge distribution is enclosed by the Gaussian surface. This leads to different expressions for the electric field depending on the position with respect to the cylinder's radius, and it highlights how geometry plays a crucial role in determining electrostatic effects.
Linear Charge Density
Linear charge density, denoted as \(\lambda\), is a measure of the amount of charge per unit length along a line or linear structure, like a wire or the axis of a cylindrical charge distribution. It is used to simplify the calculation of the electric field in cases of one-dimensional charge distributions. In the exercise provided, the linear charge density is pivotal in calculating the electric field at various points inside and outside of the cylindrical charge distribution.
Understanding linear charge density is crucial when working with electric fields and forces in systems where charge is distributed uniformly along a line. It allows us to directly relate the amount of charge within a segment to the length of that segment. For the student to grasp these concepts, it's important to connect the abstract definitions, like linear charge density, to concrete examples they can visualize, such as uniformly charged rods or filaments, and then applying these concepts to solve problems.
Understanding linear charge density is crucial when working with electric fields and forces in systems where charge is distributed uniformly along a line. It allows us to directly relate the amount of charge within a segment to the length of that segment. For the student to grasp these concepts, it's important to connect the abstract definitions, like linear charge density, to concrete examples they can visualize, such as uniformly charged rods or filaments, and then applying these concepts to solve problems.