Question

The region between two concentric conducting spheres with radii \(a\) and \(b\) is filled with a conducting material with resistivity \(\rho\). (a) Show that the resistance between the spheres is given by $$R = {\rho\over4\pi} ({1\over a}- {1\over b})$$(b) Derive an expression for the current density as a function of radius, in terms of the potential difference \(V_{ab}\) between the spheres. (c) Show that the result in part (a) reduces to Eq. (25.10) when the separation \(L = b - a\) between the spheres is small.

Short answer

Resistance is \(R = \frac{\rho}{4\pi} \left( \frac{1}{a} - \frac{1}{b} \right)\); current density \(J(r) = \frac{V_{ab}}{\rho} \frac{1}{r^2}\); in the limit, it matches the parallel plate model.

Step-by-step solution

Understand the Geometry

We have two concentric spheres with radii \(a\) and \(b\), and we need to find the resistance between them, where the space between the spheres is filled with a conducting material of resistivity \(\rho\).

Derive Resistance Formula

Consider a thin spherical shell of radius \(r\) and thickness \(dr\) between the spheres. The resistance of this shell is given by \(dR = \frac{\rho \, dr}{4 \pi r^2}\). Integrating from \(a\) to \(b\), the total resistance \(R\) is \[ R = \int_{a}^{b} \frac{\rho}{4 \pi r^2} \, dr = \frac{\rho}{4 \pi} \left( \frac{1}{a} - \frac{1}{b} \right). \]

Derive Current Density Expression

The potential difference \(V_{ab}\) drives the current through the spherical shell. Ohm's law in differential form gives \(E = -abla V\). At radius \(r\), \(E = \frac{V_{ab}}{b - a} \frac{1}{r^2}\), leading to the current density \(J\) as \[ J(r) = \frac{V_{ab}}{\rho} \frac{1}{r^2}. \]

Verify Limiting Case

When the separation \(L = b - a\) is small, the spheres become nearly parallel plates. The resistance formula \(R = \frac{\rho}{4 \pi} \left( \frac{1}{a} - \frac{1}{b} \right)\) approaches the parallel plate formula: \( R = \frac{\rho L}{A} \), where \(A\) is the surface area, \(A = 4 \pi a^2\), confirming the form is similar as \(b\) approaches \(a\).

Key concepts

Concentric Spheres

In the realm of electrical engineering and physics, concentric spheres are two or more spheres that share a common center but differ in size. When solving problems involving concentric spheres, imagine the spheres as shells around one another. The region between these inner and outer spheres can be filled with different materials, including conducting materials, as specified in the original exercise.

Understanding the geometry of concentric spheres is crucial when calculating various electrical properties, such as resistance or capacitance. The spherical symmetry simplifies potential field calculations, allowing us to slice the area into infinitesimal layers that provide an easier path for integration. This symmetry is specifically useful in the context of resistance, as seen in the derivation using a geometry-focused approach.

Resistivity

Resistivity is an intrinsic property of materials that quantifies how strongly a given material opposes the flow of electric current. It is denoted by the symbol \( \rho \) and typically measured in ohm-meters (Ω·m). In the context of the exercise, the material filling the space between the concentric spheres possesses a specific resistivity \( \rho \).

Knowing the resistivity of a material allows you to predict and calculate the behavior of electrical systems in which that material is present. For any conducting material, the resistance \( R \), which indicates how difficult it is for an electric current to pass through, is determined by its resistivity \( \rho \), the material length, and its cross-sectional area.

• The higher the resistivity, the higher the resistance for a given material.
• In the formula derived in the exercise, \( R = \frac{\rho}{4\pi} \left( \frac{1}{a} - \frac{1}{b} \right) \), it is the resistivity that scales the resistance between the spheres directly proportional to its value.

Current Density

Current density, denoted by \( J \), is an important property in electromagnetism that describes the amount of electrical current flowing per unit area through a particular point in space. It is measured in amperes per square meter (A/m²).

For the concentric spheres scenario, imagine current as a flow of water through a pipe. Current density gives you an idea of how densely "packed" this flow is, within the material's microscopic framework. In this exercise, the formula for current density is expressed as \( J(r) = \frac{V_{ab}}{\rho} \frac{1}{r^2} \), where negative signs indicate directional conventions associated with potential fields.

• The current density depends on the potential difference and resistivity.
• It inversely depends on the square of the radius, \( r \), indicating how it diminishes as we move outward from the center sphere.

Potential Difference

Potential difference, often referred to as voltage, indicates the amount of energy required to move a unit charge from one point to another within an electric field. Represented as \( V \), it is measured in volts (V).

In the example with concentric spheres, the potential difference \( V_{ab} \) is critical in determining how the electric field progresses through the conductive material. The potential difference is responsible for "pushing" the charged particles through the conductor.

• It's often the driving "force" behind current flow: akin to the water pressure that moves water through a hose.
• According to Ohm's law, the potential difference across a conductor is directly proportional to the current through it, revealing its central role in calculating current density \( J \) and thus resistance \( R \).