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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 \(R=\frac{\rho}{4 \pi}\left(\frac{1}{a}-\frac{1}{b}\right)\) (b) Derive an expression for the current density as a function of radius, in terms of the potential difference \(V_{a b}\) between the spheres.

Short Answer

Expert verified
Resistance between the spheres is \( R=\frac{\rho}{4\pi}\left(\frac{1}{a}-\frac{1}{b}\right) \). Current density \( J(r) = \frac{V_{ab}}{R \cdot 4\pi r^2} \).

Step by step solution

01

Determine the Current Flow and Geometry

Consider a small cylindrical shell of thickness \(dr\) at radius \(r\), with height equal to the separation between the spheres. The cross-sectional area for current flow is the surface area of the shell, \(A = 4\pi r^2\).
02

Apply Ohm's Law for Differential Resistance

The differential resistance \(dR\) is given by \(dR = \frac{\rho}{A} dr = \frac{\rho}{4\pi r^2} dr\). This relates the infinitesimal resistance of the shell to its geometry.
03

Integrate the Differential Resistance

To find the total resistance, integrate \(dR\) from \(r = a\) to \(r = b\): \[ R = \int_a^b \frac{\rho}{4\pi r^2} dr = \frac{\rho}{4\pi} \left[ -\frac{1}{r} \right]_a^b = \frac{\rho}{4 \pi} \left( \frac{1}{a} - \frac{1}{b} \right) \]
04

Expressing Potential Difference in Spherically Symmetric Case

The potential difference \( V_{ab} \) between the spheres is related to the resistance \( R \) and the current \( I \) through Ohm's law: \( V_{ab} = IR \).
05

Derive the Expression for Current Density

The current density \( J(r) \) is given by \( J = \frac{I}{A} \). Therefore, using the area of a sphere: \[ J(r) = \frac{I}{4\pi r^2} \] Using \( V_{ab} = IR \), substitute \(I = \frac{V_{ab}}{R}\) to get: \[ J(r) = \frac{V_{ab}}{R \cdot 4\pi r^2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Concentric Spheres
The concept of concentric spheres involves two spherical surfaces that share the same center but differ in radius. Imagine two spheres, one nested inside the other, like layers of an onion. In this scenario, the spheres have radii \(a\) and \(b\) where \(a < b\). The space in between these spheres can be used to study various physical properties, such as electrical resistance. This model is valuable in fields such as physics and engineering where symmetrical properties simplify complex calculations. By exploring the interactions between these spheres, insights into current flow and resistance in three-dimensional spaces can be gleaned.
Ohm's Law
Ohm's Law is a fundamental concept in electrical circuits, describing the relationship between voltage, current, and resistance. The law states that the current \(I\) through a conductor between two points is directly proportional to the voltage \(V\) across the two points and inversely proportional to the resistance \(R\) of the conductor: \[ V = IR \]In the context of concentric spheres, Ohm's Law helps us understand how the potential difference between the spheres affects the current flowing through the material filling the space between them. Using the derived expression for resistance, it helps in calculating the potential difference needed to drive a certain amount of current between the spheres.
Current Density
The current density \(J\) is an important concept that describes the amount of current flowing per unit area of a cross section. In our scenario, it is more insightful to consider how current is distributed over the spherical surface.
  • The formula for current density is given by: \[ J = \frac{I}{A} \]
  • For spheres, the cross-sectional area \(A\) depends on the radius, \(A = 4\pi r^2\).
  • Expressing \(J\) as a function of radius, we find: \[ J(r) = \frac{I}{4\pi r^2} \]
By understanding current density, one can infer how efficiently the current spreads over a spherical surface.
Potential Difference
The potential difference (also known as voltage) is a measure of the electrical potential energy per unit charge between two points in a circuit. In the scenario of concentric spheres, the potential difference \(V_{ab}\) is essential for determining how much energy is needed to drive a current through the conducting material between the spheres. According to Ohm’s Law, this relationship is depicted as: \[ V_{ab} = IR \]This equation shows that the potential difference is directly proportional to the current \(I\) and the total electrical resistance \(R\). Understanding potential difference allows us to design circuits where energy flows optimally between differing potentials.
Spherical Geometry
Spherical geometry is vital in understanding the physical layout of concentric spheres. Unlike planar geometry, spherical geometry involves calculations based on curved surfaces.
  • The surface area of a sphere said to be the cross-section for current flow in this model is calculated as \(4\pi r^2\).
  • This characteristic aids in the integration of resistance over spherical shells and helps to derive the expression for current density.
Mastering spherical geometry allows for more accurate modeling of scenarios where curvature plays a pivotal role, such as in electromagnetic fields or when dealing with spherical capacitors.
Resistivity
Resistivity \(\rho\) is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current. It is an inherent characteristic of the material filling the space between the concentric spheres.
  • The resistance \(R\) is directly proportional to the resistivity, meaning materials with high resistivity contribute more to resistance.
  • The formula for resistance in this setup is: \[ R = \frac{\rho}{4\pi} \left( \frac{1}{a} - \frac{1}{b} \right) \]
Understanding resistivity is crucial in selecting materials for specific applications to control the flow of electricity efficiently between conducting surfaces.

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Most popular questions from this chapter

Consider a resistor with length \(L\), uniform cross-sectional area \(A\), and uniform resistivity \(\rho\) that is carrying a current with uniform current density \(J .\) Find the electrical power dissipated per unit volume, \(p .\) Express your result in terms of (a) \(E\) and \(J ;\) (b) \(J\) and \(\rho\); (c) \(E\) and \(\rho\).

A 1.50-m cylindrical rod of diameter \(0.500 \mathrm{~cm}\) is connected to a power supply that maintains a constant potential difference of \(15.0 \mathrm{~V}\) across its ends, while an ammeter measures the current through it. You observe that at room temperature \(\left(20.0^{\circ} \mathrm{C}\right.\) ) the ammeter reads \(18 \mathrm{~A}\), while at \(90.0^{\circ} \mathrm{C}\) it reads \(16.5 \mathrm{~A}\). You can ignore any thermal expansion of the rod. Find (a) the resistivity at \(20.0^{\circ} \mathrm{C}\) and (b) the temperature coefficient of resistivity at \(20^{\circ} \mathrm{C}\) for the material of the rod.

A conductor of resistivity \(\rho\) is placed in an electric field \(E\) which produces current in it. The heat energy dissipated per second in a unit volume is - ..............

The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours \((\mathrm{A} \cdot \mathrm{h}), \mathrm{A}\) 50-A - h battery can supply a current of \(50 \mathrm{~A}\) for \(1.0 \mathrm{~h}\), or \(25 \mathrm{~A}\) for \(2.0 \mathrm{~h}\), and so on. (a) What total energy can be supplied by a \(12-\mathrm{V}\), \(60-\mathrm{A} \cdot \mathrm{h}\) battery if its internal resistance is negligible? (b) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? the density of gasoline is \(\left.900 \mathrm{~kg} / \mathrm{m}^{3} .\right)\) (c) If a generator with an average electrical power output of \(0.45 \mathrm{~kW}\) is connected to the battery, how much time will be required for it to charge the battery fully? Latent heat of combustion of gasoline \(=46 \times 10^{6} \mathrm{~J} / \mathrm{kg}\).

An electrical conductor designed to carry large currents has a circular cross section \(2.50 \mathrm{~mm}\) in diameter and is \(11 \mathrm{~m}\) long. The resistance between its ends is \(0.14 \Omega\). (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is \(1.4 \mathrm{~V} / \mathrm{m}\), what is the total current? (c) If the material has \(8.5 \times 10^{28}\) free electrons per cubic meter, find the average drift speed under the conditions of part (b).

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