Problem 39
Question
A long coaxial cable consists of an inner cylindrical conductor with radius \(a\) and an outer coaxial cylinder with inner radius \(b\) and outer radius \(c\). The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length \(\lambda\). Calculate the electric field (a) at any point between the cylinders a distance \(r\) from the axis and (b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance \(r\) from the axis of the cable, from \(r = 0\) to \(r = 2c\). (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.
Step-by-Step Solution
Verified Answer
(a) \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\) for \(a < r < b\); (b) same expression for \(r > c\); (c) plot shows piecewise function; (d) \(\lambda_{inner} = -\lambda, \lambda_{outer} = \lambda\).
1Step 1: Analyze the Electric Field Between the Cylinders
To find the electric field between the two cylinders, where radius \(a < r < b\), use Gauss's Law. Over a cylindrical Gaussian surface of radius \(r\) and length \(L\), the enclosed charge is \(\lambda L\). By symmetry, the electric field \(E\) is radial and constant over the cylindrical surface.Gauss's Law states: \[\oint E \, dA = \frac{q_{enc}}{\varepsilon_0}\]The left side becomes:\[E (2 \pi r L) = \frac{\lambda L}{\varepsilon_0}\]Solving for \(E\), the electric field at a distance \(r\) is:\[E = \frac{\lambda}{2\pi \varepsilon_0 r}\].
2Step 2: Calculate the Electric Field Outside the Outer Cylinder
For points outside the outer cylinder \(r > c\), consider the entire charge of the inner wire as if it were concentrated at the axis since the outer cylinder has no net charge. Thus, using Gauss's Law again:\[E (2\pi r L) = \frac{\lambda L}{\varepsilon_0}\]Solving for \(E\), the electric field is:\[E = \frac{\lambda}{2\pi \varepsilon_0 r}\].Interestingly, this result is the same form as the field between the cylinders, due to no additional enclosed charges beyond \(b\).
3Step 3: Graph the Electric Field as a Function of Distance
The electric field is zero for \(r < a\). For \(a < r < b\), \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\). For \(b < r < c\), \(E = 0\) since there is no charge inside this region, due to the neutrality of the outer cylinder. For \(r > c\), again \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\).Thus, the graph would rise inversely with \(r\) from zero at \(r = a\), drop to zero for \(b < r < c\), and rise inversely again for \(r > c\).
4Step 4: Determine the Charge on the Cylinders
Since the outer cylinder has no net charge but the inner cylinder has charge \(\lambda\), the outer cylinder must have equal and opposite charge on its inner surface (\(-\lambda\)) to maintain neutrality. Thus, the charge per unit length on the inner surface of the outer cylinder is \(-\lambda\), and on the outer surface, it is \(\lambda\) due to induced effects but overall remains zero net charge.
Key Concepts
Electric FieldCylindrical SymmetryCoaxial CableCharge Distribution
Electric Field
The electric field is a fundamental concept in physics, describing how electric charges interact in space. In the case of a coaxial cable, the electric field is specifically dealt with in cylindrical geometry. Gauss's Law, a cornerstone of electromagnetism, helps calculate the electric field in regions around charged objects. For our coaxial system:
- Inside the inner cylinder (\( r < a \)) - the field is zero due to the symmetry and lack of enclosed charge.
- Between the cylinders (\( a < r < b \)) - Gauss's Law provides that the field \( E \) becomes \( rac{\lambda}{2 \pi \varepsilon_0 r} \). Here, \( \lambda \) represents the linear charge density, and \( \varepsilon_0 \) is the permittivity of free space.
- Outside the outer cylinder (\( r > c \)) - the same formula applies because the outer cylinder has no net charge and does not contribute to the field beyond itself.
Cylindrical Symmetry
Cylindrical symmetry is a type of symmetry where the physical properties of a system are invariant around a central axis. In the context of a coaxial cable, this means the electric field depends only on the distance \( r \) from the axis, and not on the angle or direction around it.
For calculations, cylindrical symmetry simplifies problems because the same formula can apply over every part of the cylindrical surface at the same radial distance:
For calculations, cylindrical symmetry simplifies problems because the same formula can apply over every part of the cylindrical surface at the same radial distance:
- The electric field is uniform along any circular ring centered on the axis.
- This uniformity arises due to the symmetric charge distribution along the length of the cable.
- Gauss's Law leverages this symmetry by using a cylindrical Gaussian surface.
Coaxial Cable
A coaxial cable is a cable with two cylindrical conductors. The primary uses are for transmitting electrical signals with minimal interference. These cables consist of:
- An inner conductor, which carries the signal.
- An insulating layer, separating the inner conductor from the outer cylindrical conductor.
- An outer cylinder, which can shield the internal signal from external electrical noise.
- We consider not just the design but how different electrical charges distribute across the surfaces.
- Their cylindrical symmetry and insulation are key for understanding electromagnetic interactions within the cable.
Charge Distribution
In a coaxial cable, charge distribution affects neighboring electric fields. Charges distribute over the conductors in response to external electric influences and the conductor's constraints.
- The inner wire typically has a charge density \( \lambda \), causing electric fields in surrounding spaces.
- The outer cylinder is designed to have no net charge for shielding purposes. Despite having no net charge, it will redistribute charges on its surfaces due to induced effects.
- The inner surface of the outer cylinder holds charge \(-\lambda\) as it negates the inner wire's field internally.
- Induction ensures the outer surface may have a positive charge \( \lambda \), counterbalancing the inner surface, and maintaining neutrality externally.
Other exercises in this chapter
Problem 36
In a region of space there is an electric field \(\overrightarrow{E}\) that is in the z-direction and that has magnitude \(E =\) [964 N/(C \(\cdot\) m)]\(x\). F
View solution Problem 38
A long line carrying a uniform linear charge density +50.0 \(\mu\)C/m runs parallel to and 10.0 cm from the surface of a large, flat plastic sheet that has a un
View solution Problem 40
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 con
View solution Problem 41
A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho\). (a) Derive the expr
View solution