Question

A certain spherically symmetric charge configuration in free space produces an electric field given in spherical coordinates by $$ \mathbf{E}(r)=\left\\{\begin{array}{ll} \left(\rho_{0} r^{2}\right) /\left(100 \epsilon_{0}\right) \mathbf{a}_{r} \mathrm{~V} / \mathrm{m} & (r \leq 10) \\ \left(100 \rho_{0}\right) /\left(\epsilon_{0} r^{2}\right) \mathrm{a}_{r} \mathrm{~V} / \mathrm{m} & (r \geq 10) \end{array}\right. $$ where \(\rho_{0}\) is a constant. (a) Find the charge density as a function of position. (b) Find the absolute potential as a function of position in the two regions, \(r \leq 10\) and \(r \geq 10 .(c)\) Check your result of part \(b\) by using the gradient. (d) Find the stored energy in the charge by an integral of the form of Eq. (43). (e) Find the stored energy in the field by an integral of the form of Eq. (45).

Short answer

#tag_title# (Find absolute potential from given electric field) #tag_content# To find the absolute potential function, we will integrate the electric field over the radial distance for both regions. First, let's recall the definition of the potential difference: $$ \Delta V = -\int_\mathbf{A}^\mathbf{B} \mathbf{E} \cdot \mathrm{d} \boldsymbol{l} $$ For \(r \leq 10\), we have: $$ V(r) - V(0) = -\int_{0}^{r} \frac{\rho_0 r^2}{100\epsilon_0} dr $$ Similarly, for \(r \geq 10\), we have: $$ V(r) - V(10) = -\int_{10}^{r} \frac{\rho_0 (20-r)}{10\epsilon_0} dr $$ Now we can find the potential function for both regions by integrating and applying appropriate boundary conditions. #(c) Checking the Results by Taking the Gradient# #tag_title# (Check electric field consistency with potential) #tag_content# To check if our results in part (b) are correct, we can take the gradient of the obtained potential functions for both regions and compare with the given electric fields. For \(r \leq 10\), we have: $$ \nabla V_{r \leq 10}(r) = -\mathbf{E}_{r \leq 10}(r) $$ Similarly, for \(r \geq 10\), we have: $$ \nabla V_{r \geq 10}(r) = -\mathbf{E}_{r \geq 10}(r) $$ If the results match, then our potential functions are indeed correct. #(d) Finding the Energy Stored in the Charge# #tag_title# (Find stored energy in the charge) #tag_content# Using equation (43), we can find the energy stored in the charge and field by integrating the charge density and potential over the entire volume: $$ W_{charge} = \frac{1}{2} \int_{all\; space} \rho(\mathbf{r}) V(\mathbf{r}) d^3\mathbf{r} $$ #(e) Finding the Energy Stored in the Electric Field# #tag_title# (Find stored energy in the electric field) #tag_content# Using equation (45), we can find the energy stored in the electric field by integrating the square of the electric field magnitude over the entire volume: $$ W_{field} = \frac{1}{2} \int_{all\; space} u(\mathbf{r}) d^3\mathbf{r} $$ where \(u(\mathbf{r}) = \frac{1}{2} \epsilon_0 E^2 (\mathbf{r})\) is the energy density associated with the electric field. By following these steps, we can calculate each of the required quantities and solve the problem.

Step-by-step solution

(Find charge enclosed from given electric field)

Using the given electric field, let's find the charge enclosed in a sphere of radius r for both regions. For \(r \leq 10\), we have: $$ \mathbf{E}(r) = \frac{\rho_0 r^2}{100\epsilon_0}\mathbf{a}_r $$ Gauss's law states that, for a closed Gaussian surface, \( \oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_{enc}}{\epsilon_0}\), where \(\mathbf{dA}\) is the differential area vector, and \(Q_{enc}\) is the charge enclosed by the surface. Inside the sphere (\(r \leq 10\)): $$ \oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_{enc}}{\epsilon_0} $$ For a spherical surface, \(\mathbf{dA} = r^2 \mathbf{a}_r d\Omega\), where \(d\Omega\) is the differential solid angle. Thus, we have: $$ \int_{0}^{r} \frac{\rho_0 r^2}{100\epsilon_0} r^2 \sin\theta\, d\theta d\phi = \frac{Q_{enc}}{\epsilon_0} $$ #(b) Finding the Absolute Potential Function# inspace

Key concepts

Electric Field

The electric field is a fundamental concept in physics. It represents the force experienced by a charged particle within a given region. In spherical coordinates, an electric field can change based on the position from the source charge. In our case, when analyzing spherical symmetry, the electric field strength \(\mathbf{E}(r)\) is provided in terms of radial distance, \(r\).
This spherical symmetry implies that at any given radial distance, the field points radially outward or inward, depending on the charge sign. The formula given in the problem, \\[ \mathbf{E}(r) = \begin{cases} \frac{\rho_0 r^2}{100 \epsilon_0} \mathbf{a}_r & \text{for } r \leq 10 \ \frac{100 \rho_0}{\epsilon_0 r^2} \mathbf{a}_r & \text{for } r \geq 10 \\end{cases} \]shows that for different regions of radial distance \((r \leq 10\) and \(r \geq 10)\), the dependency of the electric field on \(r\) shifts from a quadratic relation to an inverse square relation. This change illustrates that as you move away from the center in a spherically symmetric setup, the influence of the field decreases.

Gauss's Law

Gauss's Law provides a significant link between electric fields and charge distributions. It states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space, represented mathematically as:\[\oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_{enc}}{\epsilon_0}\]In simple terms, Gauss's Law helps in calculating the enclosed charge from a known electric field.
In the exercise, applying Gauss's Law involves setting a Gaussian surface (a sphere, in this case). The symmetry here optimally exploits spherical coordinates, making it easier to integrate over the spherical surface.
As seen, for a spherical region \(r \leq 10\), Gauss's Law simplifies the integration of electric field to find the enclosed charge \(Q_{enc}\), since the electric flux depends solely on the radius \(r\). The conditions change for \(r \geq 10\), where characteristics of the electric field and the computation of charge density differ due to the inverse square relation.

Charge Density

Charge density describes how charge is distributed within a given volume, surface, or length. In our case, the focus is on the volume charge density, \(\rho_v\), which depicts how charge varies in space.
The charge density within a sphere (radially symmetric configuration) is a function of radius, \(\rho_v(r)\). It is derived from the electric field by applying Gauss’s Law and also through analyzing the sphere's volume.
For the region \(r \leq 10\), the integration of the electric field contributes directly to finding \(\rho_v\). This charge density varies as a function of \(r^2\), aligning with the expression for electric field given. With a spherically symmetric charge density, knowing \(\rho_0\) and specific radius values helps calculate the exact charge distribution.

• Inside the sphere: Charge density affects the quadratic term in the electric field equation.
• Outside the sphere: The influence of exact charge magnitude outweighs symmetry, driving the inverse square relationship.

Potential Function

The electric potential, often denoted as \(V\), represents the work done per unit charge in bringing a charge from a reference point to a specific point in the field without acceleration. It links to the electric field through:\[\mathbf{E} = -abla V\]The negative gradient of the potential function gives the electric field.
For regions \(r \leq 10\) and \(r \geq 10\), the potential function varies according to how energy is stored in the field and depends heavily on spatial configuration.
For calculating \(V\), integrate the electric field as a function of radius. This integration must respect the differential form of electric potential with respect to the radial direction:\[V(r) = - \int_r^{\infty} \mathbf{E} \cdot d\mathbf{r}\]The result gives the potential energy stored per unit charge at a given point \(r\), varying based on whether \(r \leq 10\) or \(r \geq 10\). The gradient check ensures that derived potential translates back to field calculations effectively.