Problem 37 - University Physics with Modern Physics

Question

22.37. The Coasial Cable. 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 \(r=2 c\) (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

Step-by-Step Solution

Answer

(a) Electric field is \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\) for \(a < r < b\). (b) Same formula for \(r > c\). (c) Graph shows decreasing hyperbola. (d) Charge per unit length is \(-\lambda\) on inner surface.

Step 1: Apply Gauss's Law

Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space, \(\varepsilon_0\). Mathematically, it is given by: \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \] For (a), a cylindrical Gaussian surface of radius \(r\) and length \(L\), where \( a < r < b \), is considered. The charge enclosed by this Gaussian surface is \( \lambda L \).

Step 2: Calculate the Electric Field Between Cylinders (a < r < b)

By symmetry, the electric field \( \mathbf{E} \) is radial and has the same magnitude at every point on the Gaussian surface. The total electric flux \( \oint \mathbf{E} \cdot d\mathbf{A} \) through the Gaussian surface is \(E \cdot 2\pi r L\). Substituting in Gauss's Law gives: \[ E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0} \] Solving for \(E\) yields: \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \] Thus, the electric field between the cylinders is \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\).

Step 3: Calculate the Electric Field Outside the Outer Cylinder (r > c)

For \(r > c\), consider a cylindrical Gaussian surface of radius \(r\) where \(r > c\). The total charge enclosed remains \(\lambda L\), since the outer cylinder has no net charge. The symmetry argument still applies, so \(E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0}\). Solving for \(E\) gives: \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \] Thus, the electric field outside the outer cylinder is \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\).

Step 4: Graph the Electric Field Magnitude as a Function of r

The electric field \(E\) from the solution steps is a function of \(r\): \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\). This function is relevant for both regions, \(a < r < b\) and \(r > c\), and is depicted as a decreasing hyperbola. Graph \(E(r)\) starting from near zero, peaking immediately as it moves away from \(a\), decreasing across both regions, potentially with a drop-off at \(r = b\) due to shielding.

Step 5: Determine Charge Distribution on Outer Cylinder

(d) Since the outer cylinder has no net charge, the charge per unit length induced on the inner surface of the outer cylinder is \(-\lambda\) due to charge separation caused by the inner cylinder's positive charge. On the outer surface of the outer cylinder, there is no additional charge, so it remains at \(0\).

Key Concepts

Electric Field Calculation

To understand the electric field calculation in a coaxial cable, we use Gauss's Law. This principle reveals the relationship between electric flux and the charge it encompasses. In our scenario, we are evaluating a coaxial cable that consists of an inner charged cylinder and an outer uncharged cylinder. For the electric field between these cylinders, we choose a cylindrical Gaussian surface with radius \(r\) such that \(a < r < b\). Here, \(a\) and \(b\) represent the radii of the inner and outer cylinders respectively. Applying Gauss's Law, which states \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \), helps us understand how the charge \(\lambda \) over length \(L\) influences the electric field \(E\). This setup results in the equation \(E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0}\), ultimately giving \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\).This equation indicates that the electric field is radial and inversely proportional to \(r\), highlighting its decrease as the distance increases.

Charge Distribution

Charge distribution along the coaxial cable drastically affects the electric field generated in and around the cable. The inner cylinder carries a positive charge per unit length, denoted as \(\lambda\). This singular positive charge induces a unique distribution across the cable.

• On the inner surface of the outer cylinder: Charged particles on the inner surface of the outer cylinder are influenced by the electric field generated by the inner cylinder. The charge per unit length induced here is \(-\lambda\), essentially negating the effect of the inner cylinder's charge.
• On the outer surface of the outer cylinder: Here, the charge does not change, remaining at \(0\) since the cable itself is neutral overall.

This separated charge distribution underlines how the charge on the inner cylinder is balanced by the induced charge on the inner surface of the outer cylinder, leaving the outer surface unaffected. This arrangement ensures the stable operation of the cable within different contexts.

Coaxial Cable

A coaxial cable is an essential component employed commonly in electrical and communication networks. It consists of an internal cylindrical conductor insulated from an external hollow cylindrical conductor.

• Inner Conductor: Typically, the inner conductor is a solid wire that carries a signal or charge which often requires measurement or manipulation like in Gauss's Law application.
• Outer Conductor: This segment acts both as a return path for the current and a shield to protect from external electromagnetic interference.
• Insulating Layer: Situated between the two cylinders, it ensures that the inner and outer conductors remain electrically isolated while allowing physical stability.

Due to this unique setup, coaxial cables are revered for effectively managing electromagnetic fields around them, crucial for maintaining signal integrity in communication systems. The clear segregation within its structure ensures minimal signal loss, which is vital for efficient data transmission.