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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 \((a) \mathbf{E} ;(b) \mathbf{D} ;(c) Q ;(d) C .\)

Short Answer

Expert verified
Based on the above step-by-step solution, answer the following: Q: Determine the expression for the electric field (E) as a function of ρ. A: The electric field, E(ρ), is given as: \(E(\rho) = -\frac{V_0}{d}\).

Step by step solution

01

Write Gauss's Law and E-D relation in cylindrical coordinates

In cylindrical coordinates \((\rho, \phi, z)\), Gauss's Law is given as: \(\nabla \cdot \mathbf{D} = \frac{\partial D_{\rho}}{\partial \rho} + \frac{1}{\rho}\frac{\partial D_{\phi}}{\partial \phi} + \frac{\partial D_{z}}{\partial z} = 0\) Since the electric field and displacement are z-directional only (symmetry and parallel plates), we can eliminate the other components: \(\frac{\partial D_{z}}{\partial z} = 0\) We also have the E-D relation in dielectric materials: \(\mathbf{D} = \epsilon \mathbf{E}\) where \(\epsilon = \epsilon_{0}\left(1+\rho^{2} / a^{2}\right)\) in our case.
02

Calculate E and D in z-direction using Gauss's Law and E-D relation

Using the E-D relation, we can write \(D_{z} = \epsilon_{0} E_{z} \left(1+\rho^{2} / a^{2}\right)\). Now, substitute this into Gauss's Law equation from Step 1: \(\frac{\partial (\epsilon_{0} E_{z} \left(1+\rho^{2} / a^{2}\right))}{\partial z} = 0\) Since \(\epsilon_0\) and \(a\) are constants, the equation becomes: \(\frac{\partial E_{z}}{\partial z}(1+\rho^{2} / a^{2}) + E_{z} \frac{\partial (1+\rho^{2} / a^{2})}{\partial z} = 0\) Notice that the second term will also be 0 because \(\rho\) and \(a\) are independent of \(z\). So we have: \(\frac{\partial E_{z}}{\partial z}(1+\rho^{2} / a^{2}) = 0\) which implies that \(E_z\) must be independent of z. Thus, we can denote it as \(E(\rho)\) and divide by \((1+\rho^{2} / a^{2})\): \(\frac{\partial E(\rho)}{\partial z} = 0\) Integrate both sides to find \(E(\rho)\) and use the boundary condition \(E(\rho=a) = -V_0/d\): \(E(\rho) = -\frac{V_0}{d}\) Now we can find \(D_z\) using the E-D relation: \(D_z(\rho) =-\epsilon_0 \left(1+\rho^{2} / a^{2}\right) \frac{V_0}{d} \)
03

Calculate charge (Q) on the capacitor

The charge can be calculated as an integral of \(D_z\) over the area of the plate: \(Q = -\int_{plate} D_z(\rho) dA\) Using cylindrical coordinates, we can write the integral as: \(Q = -\int_0^a \int_0^{2 \pi} \epsilon_0 \left(1+\rho^{2} / a^{2}\right) \frac{V_0}{d} \rho d\phi d\rho\) Integrate over \(\phi\) and \(\rho\): \(Q = -\frac{2 \pi \epsilon_0 V_0}{d} \int_0^a \left(1+\rho^{2} / a^{2}\right) \rho d\rho = -\frac{2 \pi \epsilon_0 V_0}{3d} a^2\)
04

Calculate capacitance (C) in the capacitor

Finally, we can find the capacitance using the formula: \(C = \frac{Q}{V_0}\) Using the value of Q from Step 3: \(C = \frac{-\frac{2 \pi \epsilon_0 V_0}{3d} a^2}{V_0} = \frac{2 \pi \epsilon_0 a^2}{3d}\)
05

Summary of results

(a) Electric field (\(\mathbf{E}\)): \(E(\rho) = -\frac{V_0}{d}\mathbf{\hat{z}}\) (b) Electric displacement (\(\mathbf{D}\)): \(D_z(\rho) =-\epsilon_0 \left(1+\rho^{2} / a^{2}\right) \frac{V_0}{d} \mathbf{\hat{z}}\) (c) Charge (Q) on capacitor: \(Q = -\frac{2 \pi \epsilon_0 V_0}{3d} a^2\) (d) Capacitance (C) of the capacitor: \(C = \frac{2 \pi \epsilon_0 a^2}{3d}\)

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle in electromagnetism stating that the total electric flux through a closed surface is proportional to the charge enclosed by the surface. It's mathematically expressed as \(abla \bullet \textbf{D} = \rho\), where \(\textbf{D}\) is the electric displacement field, and \(\rho\) is the charge density. When there's no charge within the surface, the law simplifies to \(abla \bullet \textbf{D} = 0\), which was used in our exercise for a parallel-plate capacitor with dielectric material. The symmetry of the setup allows reducing the equation to only consider the z-component of the electric displacement field, leading to the conclusion that the electric field \(E_z\) is constant across the plates of the capacitor. This key concept enables the simplification of calculations in scenarios where the symmetry can be exploited.
Electric Displacement
Electric displacement, often denoted as \(\textbf{D}\), is a vector field that appears in Maxwell's equations. It represents the combined effect of electric field \(\textbf{E}\) and the material's response to this field, characterized by the material's permittivity \(epsilon\). In the presence of a dielectric, the electric displacement is given by the relation \(\textbf{D} = epsilon \textbf{E}\). For our case, the radial dependence of the dielectric's permittivity \(epsilon(\rho) = epsilon_{0}(1 + \frac{\rho^2}{a^2})\), which adds complexity to the electric displacement calculation. The importance of \(\textbf{D}\) becomes clear when calculating the charge distribution on the capacitor's plates, as it incorporates both the effect of electric field and the modifying influence of the dielectric.
Capacitance Calculation
Capacitance is a measure of a capacitor's ability to store charge per unit voltage and is a fundamental concept in circuits. It is given by the formula \(C = \frac{Q}{V_0}\), where \(Q\) is the charge stored, and \(V_0\) is the applied voltage. In our exercise, the capacitance calculation involves integrating the electric displacement over the area of the plate to find the charge \(Q\), followed by dividing this by the voltage \(V_0\) across the capacitor. As the plates are circular and the dielectric has radial permittivity, the integration takes into account the spatial variance to accurately determine the charge. The final expression for capacitance of\(C = \frac{2 pi epsilon_0a^2}{3d}\), where \(a\) is the radius of the plates and \(d\) is the distance between them, encapsulates all the contributing factors and provides a clear understanding of how geometry and material properties affect capacitance.

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Most popular questions from this chapter

Consider an arrangement of two isolated conducting surfaces of any shape that form a capacitor. Use the definitions of capacitance (Eq. (2) in this chapter) and resistance (Eq. (14) in Chapter 5) to show that when the region between the conductors is filled with either conductive material (conductivity \(\sigma\) ) or a perfect dielectric (permittivity \(\epsilon\) ), the resulting resistance and capacitance of the structures are related through the simple formula \(R C=\epsilon / \sigma .\) What basic properties must be true about both the dielectric and the conducting medium for this condition to hold for certain?

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.

The functions \(V_{1}(\rho, \phi, z)\) and \(V_{2}(\rho, \phi, z)\) both satisfy Laplace's equation in the region \(a<\rho

A coaxial cable has conductor dimensions of \(a=1.0 \mathrm{~mm}\) and \(b=2.7 \mathrm{~mm}\). The inner conductor is supported by dielectric spacers \(\left(\epsilon_{r}=5\right)\) in the form of washers with a hole radius of \(1 \mathrm{~mm}\) and an outer radius of \(2.7 \mathrm{~mm}\), and with a thickness of \(3.0 \mathrm{~mm}\). The spacers are located every \(2 \mathrm{~cm}\) down the cable. ( \(a\) ) By what factor do the spacers increase the capacitance per unit length? \((b)\) If \(100 \mathrm{~V}\) is maintained across the cable, find \(\mathbf{E}\) at all points.

Two conducting spherical shells have radii \(a=3 \mathrm{~cm}\) and \(b=6 \mathrm{~cm} .\) The interior is a perfect dielectric for which \(\epsilon_{r}=8\). (a) Find \(C .(b)\) A portion of the dielectric is now removed so that \(\epsilon_{r}=1.0,0<\phi<\pi / 2\), and \(\epsilon_{r}=8\), \(\pi / 2<\phi<2 \pi\). Again find \(C\).

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