Question

In a certain medium, the electric potential is given by $$ V(x)=\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{-a x}\right) $$ where \(\rho_{0}\) and \(a\) are constants. ( \(a\) ) Find the electric field intensity, E. \((b)\) Find the potential difference between the points \(x=d\) and \(x=0 .(c)\) If the medium permittivity is given by \(\epsilon(x)=\epsilon_{0} e^{a x}\), find the electric flux density, \(\mathbf{D}\), and the volume charge density, \(\rho_{v}\), in the region. ( \(d\) ) Find the stored energy in the region \((0

Short answer

a) Calculate the electric field intensity (E) using the given electric potential function (V). E(x) = ___________________ b) Find the potential difference between the points x=d and x=0. V(d) - V(0) = ___________________ c) Determine the electric flux density (D) and the volume charge density (ρv) in the region. D(x) = ___________________ ρv = ___________________ d) Calculate the stored energy in the region (0<x<d), (0<y<1), (0<z<1). W_stored = ___________________

Step-by-step solution

Part (a): Find the electric field intensity, E.

Since the electric potential is given by the function V(x), we can find the electric field intensity using the negative of the derivative of V(x) with respect to x: $$ \mathbf{E}(x)=-\frac{dV}{dx} $$ Now, differentiate the given V(x) function: $$ \mathbf{E}(x)=-\frac{d}{dx}\left(\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{-a x}\right)\right) $$ Find the derivative and simplify: $$ \mathbf{E}(x)=-\frac{\rho_{0}}{a \epsilon_{0}}\left(-a e^{-a x}\right) = \frac{\rho_{0}}{\epsilon_{0}}e^{-a x} $$

Part (b): Find the potential difference between the points x=d and x=0.

To find the potential difference between two points, we can subtract the potential at the first point from the potential at the second point: $$ V(x=d)-V(x=0)=\left(\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{-a d}\right)\right) - \left(\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{0}\right)\right) $$ Now simplify the expression to get: $$ V(d)-V(0)=\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{-a d}-1+1\right)=\frac{\rho_{0}(1-e^{-ad})}{a \epsilon_{0}} $$

Part (c): Find the electric flux density, D, and the volume charge density, ρv, in the region.

The electric flux density can be found using Gauss's law in differential form: $$ \mathbf{D}(x)=\epsilon(x)\mathbf{E}(x) $$ Using the given values of ε(x) and E(x) from the problem and substituting them into the equation, we have: $$ \mathbf{D}(x)=\epsilon_{0}e^{ax}\left(\frac{\rho_{0}}{\epsilon_{0}}e^{-a x}\right)=\rho_0 $$ As the electric flux density D is constant, the volume charge density ρv can also be found using the relationship between D and ρv: $$ \nabla \cdot \mathbf{D} = \rho_v $$ Here, since D is constant and does not depend on position x, the divergence of D is 0. Therefore, the volume charge density ρv is: $$ \rho_v = 0 $$

Part (d): Find the stored energy in the region (0

The stored energy density in a dielectric medium is given by: $$ u=\frac{1}{2}\mathbf{D}(x) \cdot \mathbf{E}(x) $$ Substitute the values of D(x) and E(x) in the equation from previous parts: $$ u=\frac{1}{2}\rho_0 \left(\frac{\rho_{0}}{\epsilon_{0}}e^{-a x}\right)=\frac{\rho_0^2}{2\epsilon_0}e^{-a x} $$ To find the stored energy in the region, integrate the energy density over the volume of the region: $$ W_{stored}=\int_{0}^{d}\int_{0}^{1}\int_{0}^{1}u \, dy \, dz \, dx $$ Perform the integration with respect to y and z first: $$ W_{stored}=\int_{0}^{d}\left(\frac{\rho_0^2}{2\epsilon_0}e^{-a x}\right) dx $$ Now integrate with respect to x and solve: $$ W_{stored}=-\frac{\rho_0^2}{2a\epsilon_0}\left[e^{-ad}-1\right] $$

Key concepts

Electric Potential

Understanding the concept of electric potential is key in electromagnetism. Electric potential, often denoted as \( V \), represents the potential energy per unit charge at a point in space due to electric forces. Think of it as the "pressure" in a stream of charges just as water pressure exists in a pipeline. The units of electric potential are volts (V). This scalar quantity can help us understand how much work is needed to move a charge within an electric field.
The potential difference between two points \( A \) and \( B \) is often of interest, as it tells how much energy is gained or lost by a charge moving from one point to another. Mathematically, when you know the electric potential \( V(x) \) at a location \( x \), you can derive the electric field \( \mathbf{E}(x) \) by taking the negative derivative of \( V(x) \), showing how the energy landscape changes in space.

• Key points for electric potential: Directly relates to the work done in moving a charge.
• Measured in volts (V).
• Vital in calculations for electric field \( \mathbf{E} \).

Electric Flux Density

Electric flux density, represented by \( \mathbf{D} \), is a concept that quantifies the distribution of electric field lines in a given area. It tells us about the "flow" of an electric field through a material. The unit for electric flux density is coulombs per square meter (C/m²).
Electric flux density is directly related to both the electric field \( \mathbf{E} \) and the material's permittivity \( \epsilon \). Essentially, \( \mathbf{D} \) helps link the effects of electric fields within different materials, being crucial to determining how much electric field is "pressed" through a particular medium. This relationship, given by \( \mathbf{D}(x) = \epsilon(x)\mathbf{E}(x) \), shows how a material's permittivity impacts the field's interaction with the medium.

• \( \mathbf{D} \) depends on permittivity \( \epsilon \) and electric field \( \mathbf{E} \).
• Represents "flow" of electric field lines.
• Expressed in C/m².

Volume Charge Density

In electromagnetism, volume charge density \( \rho_v \) describes how much charge exists in a specific volume. It's essentially the charge "crowd" within a given area, leading to electric phenomena within that space. It's measured in coulombs per cubic meter (C/m³).
Volume charge density is found using the relationship with electric displacement, given by \( abla \cdot \mathbf{D} = \rho_v \). If \( \mathbf{D} \) is constant, such as in a homogeneous material, the divergence is zero, indicating no charge resides in that volume. This holds significant implications when analyzing systems of charges and is pivotal in Gauss's law applications.

• Denotes charge in a volume (C/m³).
• Central to calculating electric fields in volumes.
• Linked with electric displacement \( \mathbf{D} \).

Stored Energy in a Dielectric Medium

Stored energy within a dielectric medium can be thought of as the electrical "fuel" that powers devices and systems. When an electric field is applied to a dielectric material, part of the electrical energy is stored within the material itself. This energy is stored in the form of potential energy as the material's molecules polarize.
The energy density \( u \), expressed as \( u = \frac{1}{2}\mathbf{D}(x) \cdot \mathbf{E}(x) \), helps determine how much energy per unit volume is stored. The total energy stored is found by integrating this energy density over the volume of interest. Understanding how to compute and utilize stored energy is vital for designing efficient electronic systems, including capacitors and other energy storage components.

• Energy arises due to polarization of dielectric materials.
• Calculated using \( u \) and integrated over volume.
• Essential for assessing storage capacities in materials.