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Chapter 6: Problem 42

The hemisphere \(0

Short Answer

Expert verified
To calculate the net resistance between the two perfect conductors in a hemisphere with a conical region, we first integrated over the hemisphere excluding the conical region and then subtracted the resistance of the conical region. The final expression for the net resistance was found to be \(R_\text{net} = \frac{a^2}{2\sigma}\left[\frac{\pi}{2} - \alpha\right]\), where "a" is the radius of the hemisphere, \(\sigma\) is the conductivity, and \(\alpha\) is the angle of the conical region.

Step by step solution

01

Set up the problem

First, let's consider the hemisphere and the conical region within it. The potential difference between those two conductive regions is going to be constant, because they are at different distances from the plane on which they rest. We can rewrite the resistance equation as \(R = \frac{V}{I} = \int\limits_{r = 0}^{r = a}\int\limits_{\theta = 0}^{\theta = \pi/2} \frac{drd\theta}{\sigma}\), where \(\sigma\) is the electrical conductivity of the material.
02

Integrate over the hemisphere

We will first calculate the resistance over the hemisphere, excluding the conical region. The integral over the hemisphere is: \(R = \int\limits_{\theta = 0}^{\theta = \pi/2} \int\limits_{r = 0}^{r = a} \frac{drd\theta}{\sigma}\) We can easily integrate this expression. The result is: \(R = \frac{a^2}{2\sigma}[\theta]_{0}^{\pi/2}\).
03

Calculate the resistance of the conical region

Now, we need to subtract the resistance of the conical region from the total resistance calculated above. The integral over the conical region (\(\theta < \alpha\)) is: \(R_\alpha = \int\limits_{\theta = 0}^{\theta = \alpha} \int\limits_{r = 0}^{r = a} \frac{drd\theta}{\sigma}\) Integrating this expression gives us: \(R_\alpha = \frac{a^2}{2\sigma}[\theta]_{0}^{\alpha}\).
04

Determine the net resistance

Now that we have calculated the resistances of the hemisphere and the conical region, we can now subtract the resistance of the conical region from the resistance of the hemisphere to get the net resistance: \(R_\text{net} = R - R_\alpha = \frac{a^2}{2\sigma}\left[\frac{\pi}{2} - \alpha\right]\). This is the final expression for the net resistance between the two perfect conductors. This formula allows us to find the resistance for any arbitrary angle \(\alpha\) and any radius "a" of the hemisphere.

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Key Concepts

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

Electromagnetic Theory
Electromagnetic theory encompasses the relationship between electricity and magnetism, as described by Maxwell's equations. At the heart of this theory is the concept that electric currents and magnetic fields are interconnected. For instance, a current flowing through a wire produces a magnetic field around it, and conversely, a changing magnetic field can induce an electric current in a circuit. These principles are fundamental to understanding how electrical resistance in conductive materials works.

When we discuss the resistance of materials, we refer to their ability to resist the flow of electric current. The phenomenon is intimately connected to the distribution of electric charges within the material and how they interact under an applied electric field. The resistance in a conductive material, such as in our hemisphere exercise, can be derived by considering the electric potential and the flow of current through the geometry of the object.
Electrical Conductivity
Electrical conductivity, denoted by the symbol \(\sigma\), is a measure of how well a material can allow the flow of electric current. It鈥檚 defined as the inverse of resistivity, where a higher conductivity value indicates that the material has a lower resistance to current flow. Materials with high electrical conductivity, like copper or silver, are often used to make electrical wires and components.

In the case of the hemisphere from the exercise, understanding conductivity is critical. Conductivity is not just a property of the material, but it also governs how electric fields distribute within it. Due to the difference in conductivity between the replaced conical section and the rest of the hemisphere, there is a distinct electrical resistance associated with each region, affecting how current travels between the two perfect conductors sitting on the conducting plane.
Hemisphere Conductor Integration
The process of integrating over the shape of a hemisphere to find the resistance, as in our exercise, is a classic method found in physics and engineering to solve complex problems involving continuous charge or current distributions. This integration is essentially summing up infinitesimally small resistances across the entire volume of the hemisphere to find the total resistance.

In our problem, we performed a double integral over the hemisphere鈥檚 radius and angle, which gives us a way to methodically determine the resistance throughout the hemisphere, excluding the perfectly conducting conical region. The subtracted second integral over the cone corrects for the region that does not resist electrical flow, leaving us with the net resistance. The beautiful symmetry of the hemisphere simplifies the integration, making it possible to find a formula that captures how electrical resistance changes with the shape of the conductor. This mathematical solution is not only elegant but deeply rooted in the physical properties of the materials involved and the geometric configuration of the conductors.

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Most popular questions from this chapter

Concentric conducting spheres are located at \(r=5 \mathrm{~mm}\) and \(r=20 \mathrm{~mm}\) The region between the spheres is filled with a perfect dielectric. If the inner sphere is at \(100 \mathrm{~V}\) and the outer sphere is at \(0 \mathrm{~V}(a)\) Find the location of the \(20 \mathrm{~V}\) equipotential surface. \((b)\) Find \(E_{r, \max } \cdot(c)\) Find \(\epsilon_{r}\) if the surface charge density on the inner sphere is \(1.0 \mu \mathrm{C} / \mathrm{m}^{2}\).

In free space, let \(\rho_{v}=200 \epsilon_{0} / r^{2.4}\). (a) Use Poisson's equation to find \(V(r)\) if it is assumed that \(r^{2} E_{r} \rightarrow 0\) when \(r \rightarrow 0\), and also that \(V \rightarrow 0\) as \(r \rightarrow \infty\). (b) Now find \(V(r)\) by using Gauss's law and a line integral.

A parallel-plate capacitor is made using two circular plates of radius \(a\), with the bottom plate on the \(x y\) plane, centered at the origin. The top plate is located at \(z=d\), with its center on the \(z\) axis. Potential \(V_{0}\) is on the top plate; the bottom plate is grounded. Dielectric having radially dependent permittivity fills the region between plates. The permittivity is given by \(\epsilon(\rho)=\epsilon_{0}\left(1+\rho^{2} / a^{2}\right) .\) Find \((a) V(z) ;(b) \mathbf{E} ;(c) Q ;(d) C .\) This is a reprise of Problem \(6.8\), but it starts with Laplace's equation.

A parallel-plate capacitor is filled with a nonuniform dielectric characterized by \(\epsilon_{r}=2+2 \times 10^{6} x^{2}\), where \(x\) is the distance from one plate in meters. If \(S=0.02 \mathrm{~m}^{2}\) and \(d=1 \mathrm{~mm}\), find \(C\).

A parallel-plate capacitor has plates located at \(z=0\) and \(z=d\). The region between plates is filled with a material that contains volume charge of uniform density \(\rho_{0} \mathrm{C} / \mathrm{m}^{3}\) and has permittivity \(\epsilon\). Both plates are held at ground potential. ( \(a\) ) Determine the potential field between plates. (b) Determine the electric field intensity \(\mathbf{E}\) between plates. \((c)\) Repeat parts \((a)\) and \((b)\) for the case of the plate at \(z=d\) raised to potential \(V_{0}\), with the \(z=0\) plate grounded.
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