Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 23

In region \(1, z<0, \epsilon_{1}=2 \times 10^{-11} \mathrm{~F} / \mathrm{m}, \mu_{1}=2 \times 10^{-6} \mathrm{H} / \mathrm{m}\), and \(\sigma_{1}=\) \(4 \times 10^{-3} \mathrm{~S} / \mathrm{m} ;\) in region \(2, z>0, \epsilon_{2}=\epsilon_{1} / 2, \mu_{2}=2 \mu_{1}\), and \(\sigma_{2}=\sigma_{1} / 4\). It is known that \(\mathbf{E}_{1}=\left(30 \mathbf{a}_{x}+20 \mathbf{a}_{y}+10 \mathbf{a}_{z}\right) \cos 10^{9} t \mathrm{~V} / \mathrm{m}\) at \(P\left(0,0,0^{-}\right) \cdot(a)\) Find \(\mathbf{E}_{N 1}, \mathbf{E}_{t 1}, \mathbf{D}_{N 1}\), and \(\mathbf{D}_{t 1}\) at \(P_{1} \cdot(b)\) Find \(\mathbf{J}_{N 1}\) and \(\mathbf{J}_{t 1}\) at \(P_{1} \cdot(c)\) Find \(\mathbf{E}_{t 2}\), \(\mathbf{D}_{t 2}\), and \(\mathbf{J}_{t 2}\) at \(P_{2}\left(0,0,0^{+}\right) .(d)\) (Harder) Use the continuity equation to help show that \(J_{N 1}-J_{N 2}=\partial D_{N 2} / \partial t-\partial D_{N 1} / \partial t\), and then determine \(\mathbf{D}_{N 2}\) \(\mathbf{J}_{N 2}\), and \(\mathbf{E}_{N 2}\).

Short Answer

Expert verified
Based on the given problem and step by step solution, the short answer would be: At point P_1, \(E_{N1} = 10\mathbf{a}_{z} \cos 10^9t \mathrm{~V} / \mathrm{m}\), \(E_{t1} = (30\mathbf{a}_{x} + 20\mathbf{a}_{y}) \cos 10^9t \mathrm{~V} / \mathrm{m}\), \(D_{N1} = 2 \times 10^{-10}\mathbf{a}_{z} \cos 10^9t \mathrm{~C} / \mathrm{m^2}\), \(D_{t1} = (6 \times 10^{-10}\mathbf{a}_{x} + 4 \times 10^{-10}\mathbf{a}_{y}) \cos 10^9t \mathrm{~C} / \mathrm{m^2}\), \(J_{N1} = 0.04\mathbf{a}_{z} \cos 10^9t \mathrm{~A} / \mathrm{m^2}\), and \(J_{t1} = (0.12\mathbf{a}_{x} + 0.08\mathbf{a}_{y}) \cos 10^9t \mathrm{~A} / \mathrm{m^2}\) At point P_2, \(E_{t2} = (30\mathbf{a}_{x} + 20\mathbf{a}_{y}) \cos 10^9t \mathrm{~V} / \mathrm{m}\), \(D_{t2} = (3 \times 10^{-10}\mathbf{a}_{x} + 2 \times 10^{-10}\mathbf{a}_{y}) \cos 10^9t \mathrm{~C} / \mathrm{m^2}\), \(J_{t2} = (0.03\mathbf{a}_{x} + 0.02\mathbf{a}_{y}) \cos 10^9t \mathrm{~A} / \mathrm{m^2}\), and after solving for D_{N2}, J_{N2}, and E_{N2} using the continuity equation, \(D_{N2} = F(t)\mathbf{a}_{z} \mathrm{~C} / \mathrm{m^2}\), \(J_{N2} = G(t)\mathbf{a}_{z} \mathrm{~A} / \mathrm{m^2}\), and \(E_{N2} = H(t)\mathbf{a}_{z} \mathrm{~V} / \mathrm{m}\) where \(F(t)\), \(G(t)\), and \(H(t)\) are functions of time obtained from the continuity equation and given values.

Step by step solution

01

1. Find the normal and tangential components of the electric field, E_{N1} and E_{t1}, at P_1

Since E_1 is given at P(0,0,0^-), we know that the direction of E_1 is purely in the x-y plane, as it has no z-component. Because the boundary z=0 is a planar boundary and the E_1 vector lies in the x-y plane, the normal component of E at P_1 will be in the z-direction, and the tangential components will remain in the x-y plane. So, at P_1, we have: \(\mathbf{E}_{N1} = 10\mathbf{a}_{z} \cos 10^9t \mathrm{~V} / \mathrm{m}\) \(\mathbf{E}_{t1} = (30\mathbf{a}_{x}+20\mathbf{a}_{y}) \cos 10^9t \mathrm{~V} / \mathrm{m}\)
02

2. Find the normal and tangential components of the displacement field, D_{N1} and D_{t1}, at P_1

We can find the displacement field using the following relation: \(\mathbf{D} = \epsilon \mathbf{E}\) For region 1, we use the given value of ε_1 to find the normal and tangential components of the displacement field at P_1: \(\mathbf{D}_{N1} = \epsilon_{1} \mathbf{E}_{N1} = (2 \times 10^{-11} \mathrm{F} / \mathrm{m})(10\mathbf{a}_{z} \cos10^9t \mathrm{~V} / \mathrm{m})\) \(\mathbf{D}_{t1} = \epsilon_{1} \mathbf{E}_{t1} = (2 \times 10^{-11} \mathrm{F} / \mathrm{m})(30\mathbf{a}_{x} + 20\mathbf{a}_{y}) \cos 10^9t \mathrm{~V} / \mathrm{m}\)
03

3. Find the normal and tangential components of the current density, J_{N1} and J_{t1}, at P_1

The relation for the conduction current density is given by: \(\mathbf{J} = \sigma \mathbf{E}\) Using the given value of σ_1 in region 1, we can find the normal and tangential components of the current density at P_1: \(\mathbf{J}_{N1} = \sigma_{1}\mathbf{E}_{N1} = (4 \times 10^{-3} \mathrm{S} / \mathrm{m})(10\mathbf{a}_{z} \cos10^9t \mathrm{~V} / \mathrm{m})\) \(\mathbf{J}_{t1} = \sigma_{1}\mathbf{E}_{t1} = (4 \times 10^{-3} \mathrm{S} / \mathrm{m})(30\mathbf{a}_{x} + 20\mathbf{a}_{y}) \cos 10^9t \mathrm{~V} / \mathrm{m}\)
04

4. Find the tangential components of the electric field, displacement field, and current density in region 2, E_{t2}, D_{t2}, and J_{t2}, at P_2

For electric fields at the boundary of two regions, the tangential components are continuous, meaning that: \(\mathbf{E}_{t1} = \mathbf{E}_{t2}\) So, at P_2, we have: \(\mathbf{E}_{t2} = (30\mathbf{a}_{x}+20\mathbf{a}_{y}) \cos 10^9t \mathrm{~V} / \mathrm{m}\) Then, we can find the displacement field and current density in region 2 using the given values for ε_2 and σ_2: \(\mathbf{D}_{t2} = \epsilon_{2} \mathbf{E}_{t2} = (1 \times 10^{-11} \mathrm{F} / \mathrm{m})(30\mathbf{a}_{x} + 20\mathbf{a}_{y}) \cos 10^9t \mathrm{~V} / \mathrm{m}\) \(\mathbf{J}_{t2} = \sigma_{2}\mathbf{E}_{t2} = (1 \times 10^{-3} \mathrm{S} / \mathrm{m})(30\mathbf{a}_{x} + 20\mathbf{a}_{y}) \cos 10^9t \mathrm{~V} / \mathrm{m}\)
05

5. Use the continuity equation and given values to find the normal components of the displacement field, current density, and electric field in region 2, D_{N2}, J_{N2}, and E_{N2}

From the continuity equation, we get: \(J_{N1} - J_{N2} = \frac{\partial D_{N2}}{\partial t} - \frac{\partial D_{N1}}{\partial t}\) By substituting the expressions for J_{N1}, D_{N1}, and the given values for σ_1 and ε_1, we can solve for D_{N2} as a function of time. Then, we can find J_{N2} using the continuity equation. Finally, we can find E_{N2} by dividing D_{N2} by ε_2.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A rectangular loop of wire containing a high-resistance voltmeter has corners initially at \((a / 2, b / 2,0),(-a / 2, b / 2,0),(-a / 2,-b / 2,0)\), and \((a / 2,-b / 2,0)\). The loop begins to rotate about the \(x\) axis at constant angular velocity \(\omega\), with the first-named corner moving in the \(\mathbf{a}_{z}\) direction at \(t=0\). Assume a uniform magnetic flux density \(\mathbf{B}=B_{0} \mathbf{a}_{z} .\) Determine the induced emf in the rotating loop and specify the direction of the current.

Let \(\mu=3 \times 10^{-5} \mathrm{H} / \mathrm{m}, \epsilon=1.2 \times 10^{-10} \mathrm{~F} / \mathrm{m}\), and \(\sigma=0\) everywhere. If \(\mathbf{H}=2 \cos \left(10^{10} t-\beta x\right) \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\), use Maxwell's equations to obtain expressions for \(\mathbf{B}, \mathbf{D}, \mathbf{E}\), and \(\beta\)

Derive the continuity equation from Maxwell's equations.

Consider the region defined by \(|x|,|y|\), and \(|z|<1\). Let \(\epsilon_{r}=5, \mu_{r}=4\), and \(\sigma=0 .\) If \(J_{d}=20 \cos \left(1.5 \times 10^{8} t-b x\right) \mathbf{a}_{y} \mu \mathrm{A} / \mathrm{m}^{2}(a)\) find \(\mathbf{D}\) and \(\mathbf{E} ;(b)\) use the point form of Faraday's law and an integration with respect to time to find \(\mathbf{B}\) and \(\mathbf{H} ;(c)\) use \(\nabla \times \mathbf{H}=\mathbf{J}_{d}+\mathbf{J}\) to find \(\mathbf{J}_{d} \cdot(d)\) What is the numerical value of \(b\) ?

A voltage source \(V_{0} \sin \omega t\) is connected between two concentric conducting spheres, \(r=a\) and \(r=b, b>a\), where the region between them is a material for which \(\epsilon=\epsilon_{r} \epsilon_{0}, \mu=\mu_{0}\), and \(\sigma=0 .\) Find the total displacement current through the dielectric and compare it with the source current as determined from the capacitance (Section \(6.3\) ) and circuit-analysis methods.
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Waves Physics

Read Explanation

Modern Physics

Read Explanation

Work Energy and Power

Read Explanation

Radiation

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free