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(a) Use Maxwell's first equation, \(\nabla \cdot \mathbf{D}=\rho_{v}\), to describe the variation of the electric field intensity with \(x\) in a region in which no charge density exists and in which a nonhomogeneous dielectric has a permittivity that increases exponentially with \(x\). The field has an \(x\) component only; \((b)\) repeat part \((a)\), but with a radially directed electric field (spherical coordinates), in which again \(\rho_{v}=0\), but in which the permittivity decreases exponentially with \(r\).

Short Answer

Expert verified
Answer: The expression for the variation of radially directed electric field intensity with radial distance (r) in a region with a nonhomogeneous dielectric is given by \(E_r(r) = \frac{C_2}{\epsilon(r) r^2}\), where \(C_2\) is a constant.

Step by step solution

01

Write Maxwell's first equation in terms of given variables

In this problem, we are given the equation \(\nabla \cdot \mathbf{D}=\rho_{v}\). Since no charge density exists, we have \(\rho_{v}=0\) and the equation becomes \(\nabla \cdot \mathbf{D} = 0\).
02

Express \(\mathbf{D}\) in terms of electric field intensity \(E_x\) and permittivity

For a nonhomogeneous dielectric with increasing permittivity, we have \(\mathbf{D} = \epsilon \mathbf{E}\). Since the electric field only has an \(x\) component, we can write: \(\mathbf{D} = \epsilon(x) E_x(x) \hat{x}\)
03

Calculate divergence of \(\mathbf{D}\)

Now, we need to find the divergence of \(\mathbf{D}\): \(\nabla \cdot \mathbf{D} = \frac{\partial (\epsilon(x) E_x(x))}{\partial x}\)
04

Set the divergence to zero and solve for \(E_x\)

Using the derived divergence from step 3 and the fact that \(\rho_v=0\), we now have: \(\frac{\partial (\epsilon(x) E_x(x))}{\partial x} = 0\) Integrating both sides with respect to \(x\) yields: \(\epsilon(x) E_x(x) = C_1\) We finally get the expression for the variation of electric field intensity with \(x\): \(E_x(x) = \frac{C_1}{\epsilon(x)}\) (b)
05

Repeat steps 1 and 2 for spherical coordinates

Using Maxwell's first equation in spherical coordinates, the divergence of \(\mathbf{D}\) for a radially varying electric field is given by: \(\nabla \cdot \mathbf{D} = \frac{1}{r^2} \frac{\partial (\epsilon(r) E_r(r) r^2)}{\partial r}\) Since \(\rho_{v} = 0\), the equation becomes: \(\frac{1}{r^2} \frac{\partial (\epsilon(r) E_r(r) r^2)}{\partial r} = 0\)
06

Solve the equation for \(E_r\)

Now, we need to solve this equation for \(E_r(r)\): \(\frac{\partial (\epsilon(r) E_r(r) r^2)}{\partial r} = 0\) Integrating both sides with respect to \(r\) yields: \(\epsilon(r) E_r(r) r^2 = C_2\) Finally, we get the expression for the variation of radially directed electric field intensity with \(r\): \(E_r(r) = \frac{C_2}{\epsilon(r) r^2}\)

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Key Concepts

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

Electric Field Intensity
Electric field intensity, often denoted as \( \mathbf{E} \), is a fundamental concept that describes the force per unit charge exerted on a positive test charge placed in a field.
In simple terms, it tells us how strong the electric field is at a given point.
The units are typically volts per meter (V/m).

In this exercise, we only consider the X-component of the field, denoted as \( E_x \). This means the field only has direction along the x-axis.
For a clearer understanding, think of \( E_x \) as a field pointing in one specific direction, rather than occurring in all directions like a point source would.

Connecting electric field intensity to the displacement field \( \mathbf{D} \), we have the relation \( \mathbf{D} = \epsilon \mathbf{E} \), where \( \epsilon \) is the permittivity of the medium. This relationship is crucial, as it links the material properties (via permittivity) to the electric field intensity.
Nonhomogeneous Dielectric
A nonhomogeneous dielectric is a material where the permittivity \( \epsilon \), the measure of how much electric field is permitted to "pass through," changes with position.
In other words, \( \epsilon \) is not constant but varies spatially.
This occurs often in real-world materials where composition or other factors may not be uniform throughout the material.

In this problem, the dielectric's permittivity either increases or decreases exponentially. For part (a), as you move in the x-direction, \( \epsilon(x) \) increases exponentially. This means the dielectric becomes more polarized in response to an electric field as you move in that direction.

Understanding nonhomogeneous dielectrics is key to solving problems involving varying electric fields within materials that change their response to the field over distance.
Spherical Coordinates
Spherical coordinates are used when dealing with problems where symmetry about a point is present, such as fields surrounding spherical objects.
In this coordinate system, the location of a point is given by three values: radial distance \( r \), polar angle \( \theta \), and azimuthal angle \( \phi \).
This system is particularly useful for simplifying the mathematics of three-dimensional problems.

In this exercise, we apply spherical coordinates to analyze a radially directed electric field, meaning the field lines radiate outwards like the spokes of a wheel from a central point.
This setup often occurs with problems involving spherically symmetric charge distributions or fields influenced by spherical objects.
Permittivity Variation
Permittivity variation describes how the ability of a material to allow the passage of electric field lines changes spatially.
This is crucial in materials where the electrical properties are not consistent throughout, affecting how the electric field behaves in and around the material.

In part (a) of the problem, permittivity \( \epsilon(x) \) increases exponentially with \( x \), implying that as you move, the material becomes more capable of storing electric potential energy.
In part (b), where spherical coordinates are used, permittivity \( \epsilon(r) \) decreases exponentially with radial distance \( r \), indicating a weakening capability the further you move from the center.

Grasping how permittivity varies is essential for predicting the intensity of electric fields and understanding how energy is distributed within materials. This concept is vital in designing materials and systems in electrostatics and electromagnetism.

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Most popular questions from this chapter

A radial electric field distribution in free space is given in spherical coordinates as: $$ \begin{array}{l} \mathbf{E}_{1}=\frac{r \rho_{0}}{3 \epsilon_{0}} \mathbf{a}_{r} \quad(r \leq a) \\ \mathbf{E}_{2}=\frac{\left(2 a^{3}-r^{3}\right) \rho_{0}}{3 \epsilon_{0} r^{2}} \mathbf{a}_{r} \quad(a \leq r \leq b) \\ \mathbf{E}_{3}=\frac{\left(2 a^{3}-b^{3}\right) \rho_{0}}{3 \epsilon_{0} r^{2}} \mathbf{a}_{r} \quad(r \geq b) \end{array} $$ where \(\rho_{0}, a\), and \(b\) are constants. \((a)\) Determine the volume charge density in the entire region \((0 \leq r \leq \infty)\) by the appropriate use of \(\nabla \cdot \mathbf{D}=\rho_{v} \cdot(b) \mathrm{In}\) terms of given parameters, find the total charge, \(Q\), within a sphere of radius \(r\) where \(r>b\).

Volume charge density is located as follows: \(\rho_{v}=0\) for \(\rho<1 \mathrm{~mm}\) and for \(\rho>2 \mathrm{~mm}, \rho_{v}=4 \rho \mu \mathrm{C} / \mathrm{m}^{3}\) for \(1<\rho<2 \mathrm{~mm} .\) (a) Calculate the total charge in the region \(0<\rho<\rho_{1}, 0

A uniform volume charge density of \(80 \mu \mathrm{C} / \mathrm{m}^{3}\) is present throughout the region \(8 \mathrm{~mm}10 \mathrm{~mm}\), find \(D_{r}\) at \(r=20 \mathrm{~mm}\).

Let \(\mathbf{D}=5.00 r^{2} \mathbf{a}_{r} \mathrm{mC} / \mathrm{m}^{2}\) for \(r \leq 0.08 \mathrm{~m}\) and \(\mathbf{D}=0.205 \mathrm{a}_{r} / r^{2} \mu \mathrm{C} / \mathrm{m}^{2}\) for \(r \geq 0.08 \mathrm{~m} .(a)\) Find \(\rho_{v}\) for \(r=0.06 \mathrm{~m} .(b)\) Find \(\rho_{v}\) for \(r=0.1 \mathrm{~m} .(c)\) What surface charge density could be located at \(r=0.08 \mathrm{~m}\) to cause \(\mathbf{D}=0\) for \(r>0.08 \mathrm{~m} ?\)

(a) A flux density field is given as \(\mathbf{F}_{1}=5 \mathbf{a}_{z} .\) Evaluate the outward flux of \(\mathbf{F}_{1}\) through the hemispherical surface, \(r=a, 0<\theta<\pi / 2,0<\phi<2 \pi\) (b) What simple observation would have saved a lot of work in part \(a ?\) (c) Now suppose the field is given by \(\mathbf{F}_{2}=5 z \mathbf{a}_{z} .\) Using the appropriate surface integrals, evaluate the net outward flux of \(\mathbf{F}_{2}\) through the closed surface consisting of the hemisphere of part \(a\) and its circular base in the \(x y\) plane. ( \(d\) ) Repeat part \(c\) by using the divergence theorem and an appropriate volume integral.

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