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Two identical conducting plates, each having area \(A\), are located at \(z=0\) and \(z=d\). The region between plates is filled with a material having \(z\) -dependent conductivity, \(\sigma(z)=\sigma_{0} e^{-z / d}\), where \(\sigma_{0}\) is a constant. Voltage \(V_{0}\) is applied to the plate at \(z=d ;\) the plate at \(z=0\) is at zero potential. Find, in terms of the given parameters, \((a)\) the resistance of the material; \((b)\) the total current flowing between plates; ( \(c\) ) the electric field intensity \(\mathbf{E}\) within the material.

Short Answer

Expert verified
Question: Calculate the total resistance, total current, and electric field intensity within a material with the following parameters: d = 2 cm, A = 20 cm虏, 蟽鈧 = 0.04 S/m, and voltage V鈧 = 50 V. Step 1: Calculate total resistance R = (d(1 - e)/蟽鈧A) = (2(1-2.718)/0.04*20) = (2(-1.718)/0.8) = -4.295 惟 Step 2: Calculate total current I = (蟽鈧AV鈧)/(d(1 - e)) = (0.04 * 20 * 50)/(2(1 - 2.718)) = 40/(-3.436) = -11.642 A Step 3: Calculate electric field intensity at z E(z) = (V鈧)/(d(1 - e)*e^{z/d}) = (50)/(2(1 - 2.718)*e^{z/2}) (z in cm) The total resistance within the material is -4.295 惟, the total current flowing between the plates is -11.642 A, and the electric field intensity within the material is given by E(z) = (50)/(2(1 - 2.718)*e^{z/2}) when z is measured in cm.

Step by step solution

01

Compute the resistance

To determine the resistance, we will use the relationship R = 蟻L/A, where 蟻 is resistivity, L is the distance between the plates, and A is the area of the plate. Since we are given 蟽(z), the conductivity, instead of resistivity, we will need to use the relationship 蟻(z) = 1/蟽(z). The total resistance is obtained by integrating from z=0 to z=d: R = 鈭玔蟻(z)L/Area]dz = (鈭玔1/蟽(z)]dz)/A Now substitute the expression for 蟽(z) and integrate with respect to z: R = (鈭玔1/(蟽_{0}e^{-z/d})]dz)/A
02

Evaluate the integral to find the resistance

Now, we will evaluate the integral and solve for R: R = ([-d*e^{z/d}]/(蟽_{0}A))|_{0}^{d} Evaluate the limits and simplify the expression: R = ([-d*e^{d/d} - (-d*e^{0})]/(蟽_{0}A)) R = ([-d*e^{1} + d]/(蟽_{0}A)) So, the total resistance is given by: R = (d(1 - e)/蟽_{0}A)
03

Compute the total current between the plates

To find the total current I, we can use Ohm's law V = IR. We already have the resistance R, and we are given the voltage V鈧. So, we can rewrite Ohm's law and solve for I: I = V鈧/R Substituting the given voltage and the resistance we found in step 2: I = V鈧 / (d(1 - e)/蟽_{0}A) Solve for the total current I: I = (蟽_{0}AV鈧)/(d(1 - e))
04

Compute the electric field intensity within the material

To calculate the electric field intensity E within the material, we can use the relationship between current density J and conductivity 蟽: J = 蟽E. The current flowing between the plates is perpendicular to them, so we need to find the current per unit area: J = I/A Now, substituting the expression for I from step 3 and the expression for the conductivity 蟽(z): J = (蟽_{0}V鈧)/(d(1 - e)) 蟽(z)E(z) = (蟽_{0}V鈧)/(d(1 - e)) Now, we can solve for E(z): E(z) = (V鈧)/(d(1 - e)*e^{z/d}) So, the electric field intensity within the material is given by: E(z) = (V鈧)/(d(1 - e)*e^{z/d})

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Key Concepts

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

Conductivity
Conductivity is an essential concept in electromagnetic theory as it describes how well a material can conduct electric current. The conductivity (\(\sigma\)) is the ability of a material to allow the flow of electric charge, usually measured in siemens per meter (\(S/m\)). Conductivity is the reciprocal of resistivity, which means substances with high conductivity have low resistivity.

In the exercise, conductivity varies with position according to the function \(\sigma(z) = \sigma_{0} e^{-z/d}\). Here, \(\sigma_{0}\) is a constant representing the maximum conductivity when \(z = 0\). The exponential factor \(e^{-z/d}\) indicates that conductivity decreases as \(z\) increases. This scenario is common in layered materials where the conductivity changes depending on the depth or position.
Resistance Calculation
Resistance (\(R\)) is the opposition that a material presents to the flow of electric current. According to Ohm's law, resistance can be expressed as \(R = \rho \frac{L}{A}\), where \(\rho\) is resistivity, \(L\) is the length over which current flows, and \(A\) is the cross-sectional area.

In the exercise, we start with conductance and need to use the formula for resistivity, \(\rho(z) = \frac{1}{\sigma(z)}\). Then, the total resistance between the plates can be calculated through an integral \(R = \frac{1}{A} \int_{0}^{d} \frac{1}{\sigma(z)} \, dz\). Solving this, we find that the resistance is \(R = \frac{d(1 - e)}{\sigma_{0}A}\).

Resistance calculation is crucial to understanding how materials impede the flow of electrons, and using integration helps us evaluate this for materials with varying conductivities.
Ohm's Law
Ohm鈥檚 Law is a fundamental principle in physics, describing the relationship between voltage (\(V\)), current (\(I\)), and resistance (\(R\)). Mathematically, it is expressed as \(V = IR\). Ohm's law provides a simple way to calculate the electrical parameters if any two of them are known.

In this exercise, we use Ohm's Law to find the current flowing between the plates. Since voltage \(V_{0}\) is applied across the plates and we calculated resistance \(R\), the current can be found by rearranging Ohm's law to \(I = \frac{V_{0}}{R}\). Inserting the value for resistance, we find \(I = \frac{\sigma_{0}AV_{0}}{d(1 - e)}\).
  • Voltage \(V_{0}\) is comparable to energy applied to push electrons through a material.
  • Current \(I\) reflects the flow of electric charge.
  • Resistance \(R\) opposes the flow of current.
This law allows you to see how changes in any part of the system affect the other parts.
Electric Field Intensity
Electric field intensity (\(\mathbf{E}\)) is a vector quantity that represents the force per unit charge at any point in space. It's crucial in understanding how electric forces act on charges. The electric field intensity is related to voltage and distance through the formula \(E = -\frac{dV}{dz}\).

In the material between the plates, the electrical behavior is governed by the relationship \(J = \sigma E\), where \(J\) is the current density. To find \(E\), we start from where current density equals current per area, \(J = \frac{I}{A}\). Substituting expression for current \(I\) and conductivity \(\sigma(z)\):
\[ E(z) = \frac{V_{0}}{d(1 - e) \cdot e^{z/d}} \].
Electric field intensity provides insights into how voltage applied across the plates is concentrated or dispersed within a material, affecting how charges move throughout the space.

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Most popular questions from this chapter

A large brass washer has a \(2-\mathrm{cm}\) inside diameter, a \(5-\mathrm{cm}\) outside diameter, and is \(0.5 \mathrm{~cm}\) thick. Its conductivity is \(\sigma=1.5 \times 10^{7} \mathrm{~S} / \mathrm{m} .\) The washer is cut in half along a diameter, and a voltage is applied between the two rectangular faces of one part. The resultant electric field in the interior of the half- washer is \(\mathbf{E}=(0.5 / \rho) \mathbf{a}_{\phi} \mathrm{V} / \mathrm{m}\) in cylindrical coordinates, where the \(z\) axis is the axis of the washer. (a) What potential difference exists between the two rectangular faces? ( \(b\) ) What total current is flowing? ( \(c\) ) What is the resistance between the faces?

Consider a composite material made up of two species, having number densities \(N_{1}\) and \(N_{2}\) molecules \(/ \mathrm{m}^{3}\), respectively. The two materials are uniformly mixed, yielding a total number density of \(N=N_{1}+N_{2}\). The presence of an electric field \(\mathbf{E}\) induces molecular dipole moments \(\mathbf{p}_{1}\) and \(\mathbf{p}_{2}\) within the individual species, whether mixed or not. Show that the dielectric constant of the composite material is given by \(\epsilon_{r}=f \epsilon_{r 1}+(1-f) \epsilon_{r 2}\), where \(f\) is the number fraction of species 1 dipoles in the composite, and where \(\epsilon_{r 1}\) and \(\epsilon_{r 2}\) are the dielectric constants that the unmixed species would have if each had number density \(N\).

Given the potential field \(V=100 x z /\left(x^{2}+4\right) \mathrm{V}\) in free space: \((a)\) Find \(\mathrm{D}\) at the surface \(z=0 .(b)\) Show that the \(z=0\) surface is an equipotential surface. ( \(c\) ) Assume that the \(z=0\) surface is a conductor and find the total charge on that portion of the conductor defined by \(0

Let \(\mathrm{J}=400 \sin \theta /\left(r^{2}+4\right)\) a \(_{r} \mathrm{~A} / \mathrm{m}^{2} \cdot(a)\) Find the total current flowing through that portion of the spherical surface \(r=0.8\), bounded by \(0.1 \pi<\theta<0.3 \pi, 0<\phi<2 \pi .(b)\) Find the average value of \(\mathbf{J}\) over the defined area.

Two parallel circular plates of radius \(a\) are located at \(z=0\) and \(z=d\). The top plate \((z=d)\) is raised to potential \(V_{0} ;\) the bottom plate is grounded. Between the plates is a conducting material having radial-dependent conductivity, \(\sigma(\rho)=\sigma_{0} \rho\), where \(\sigma_{0}\) is a constant. \((a)\) Find the \(\rho\) -independent electric field strength, \(\mathbf{E}\), between plates. \((b)\) Find the current density, \(\mathbf{J}\) between plates. ( \(c\) ) Find the total current, \(I\), in the structure. \((d)\) Find the resistance between plates.

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