Problem 88 - University Physics with Modern Physics

Question

In a certain region, a charge distribution exists that is spherically symmetric but nonuniform. That is, the volume charge density \(\rho(r)\) depends on the distance \(r\) from the center of the distribution but not on the spherical polar angles \(\theta\) and \(\phi .\) The electric potential \(V(r)\) due to this charge distribution is $$V(r)=\left\\{\begin{array}{l}{\frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a} \\ {0} & {\text { for } r \geq a}\end{array}\right.$$ where \(\rho_{0}\) is a constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) and \(a\) is a constant having units of meters. (a) Derive expressions for \(\vec{E}\) for the regions \(r \leq a\) and \(r \geq a .[\)Hint\(:\) Use Eq. \((23.23) .]\) Explain why \(\vec{\boldsymbol{E}}\) has only a radial component. (b) Derive an expression for \(\rho(r)\) in each of the two regions \(r \leq a\) and \(r \geq a .[\)Hint t: Use Gauss's law for two spherical shells, one of radius \(r\) and the other of radius \(r+d r .\) The charge contained in the infinitesimal spherical shell of radius \(d r\) is \(d q=4 \pi r^{2} \rho(r) d r . ](c)\) Show that the net charge contained in the volume of a sphere of radius greater than or equal to \(a\) is zero. [Hint: Integrate the expressions derived in part (b) for \(\rho(r)\) over a spherical volume of radius greater than or equal to \(a.\)) Is this result consistent with the electric field for \(r>a\) that you calculated in part (a)?

Step-by-Step Solution

Answer

For \( r \leq a \), \( \vec{E} = -\nabla V \). For \( r \geq a \), \( E = 0 \); \( \rho(r) = 0 \). Net charge for \( r \geq a \) is zero.

Step 1: Determine the electric field for r ≤ a

For a charge distribution with spherical symmetry, the electric field \( \vec{E} \) can be found using the relation \( \vec{E} = -abla V(r) \). For \( r \leq a \), the potential \( V(r) \) is given by \( V(r) = \frac{\rho_{0} a^{2}}{18 \epsilon_{0}} \left[ 1-3\left( \frac{r}{a} \right)^{2} + 2\left( \frac{r}{a} \right)^{3} \right] \). Calculate the derivative \( \frac{dV}{dr} \) and use it to find \( \vec{E} \). The electric field only has a radial component because of the symmetry of the problem.

Step 2: Calculate E for r ≥ a

For \( r \geq a \), \( V(r) = 0 \), hence the electric field \( \vec{E} = - abla V(r) = 0 \). This implies that the field is zero outside the sphere, which is consistent with a spherically symmetric charge distribution where the net charge is confined within the sphere of radius \( a \).

Step 3: Derive charge density ρ(r) for r ≤ a

Using Gauss's law, relate the electric field \( \vec{E} \) inside to the charge density \( \rho(r) \). For a spherical shell of radius \( r \leq a \), use \( abla \cdot \vec{E} = \frac{\rho(r)}{\epsilon_0} \) and the expression for \( \vec{E} \) to find \( \rho(r) \).

Step 4: Derive charge density ρ(r) for r ≥ a

As derived in Step 2, \( \vec{E} = 0 \) for \( r \geq a \). This implies \( \rho(r) = 0 \) since no electric field exists outside the charge distribution, consistent with the symmetry and Gaussian surface analysis.

Step 5: Show net charge for r ≥ a is zero

Integrate \( \rho(r) \) over a spherical volume of radius greater than or equal to \( a \). Since \( \rho(r) = 0 \) for \( r \geq a \), the integral of the charge density over this region yields zero net charge. This confirms the result that the electric field \( \vec{E} \) is zero for \( r \geq a \), validating the consistency with part (a).

Key Concepts

Electric Field Derivation

The derivation of the electric field \( \vec{E} \) is crucial in understanding how charges influence their surroundings. For a spherically symmetric charge distribution, like the one described here, the electric field primarily depends on the radial distance \( r \) from the center.

To find \( \vec{E} \), we use the relationship \( \vec{E} = - abla V(r) \), where \( V(r) \) is the electric potential. In the region \( r \leq a \), the given potential is a function of \( r \), allowing us to derive the electric field by calculating the derivative of \( V(r) \) with respect to \( r \). This results in an electric field that has only a radial component due to the symmetry of the potential function.

For \( r \geq a \), the potential \( V(r) \) is zero, indicating that the electric field is also zero outside this region. This consistent result showcases the confined nature of the charge within the specified sphere.

Spherical Symmetry

Spherical symmetry is a fundamental characteristic of our problem. It implies that all physical quantities depend only on the distance \( r \) from the center and not on other angles, such as \( \theta \) and \( \phi \).

This symmetry simplifies the study of electric fields and potentials, as it means that the solutions are naturally uniform across any surface at a constant \( r \). The radial nature of the electric field arises from this symmetry, simplifying both theoretical derivations and practical calculations.

Understanding spherical symmetry helps predict how the electric field and potential change across different regions, making problems involving spherical charge distributions more manageable.

Gauss's Law

Gauss's Law is a powerful tool for understanding electric fields, especially in cases with high symmetry, such as spherical symmetry. The law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. Mathematically, it is expressed as \( \Phi = \frac{Q}{\epsilon_0} \), where \( \Phi \) represents the electric flux and \( Q \) is the charge enclosed.

In our spherical charge distribution, we apply Gauss's Law to a spherical surface of radius \( r \). For \( r \leq a \), the law helps us relate the electric field to the charge density \( \rho(r) \) using the formula \( abla \cdot \vec{E} = \frac{\rho(r)}{\epsilon_0} \). This approach quickly reveals the form of \( \rho(r) \) and confirms the electric field behavior derived earlier.

Applying Gauss's Law to the region \( r \geq a \) further confirms that the electric field is zero, as no additional charge exists beyond this radius.

Electric Potential

The electric potential \( V(r) \) indicates the potential energy per unit charge due to the charge distribution. It provides insights into how the charge is distributed in space.

In our specified region \( r \leq a \), \( V(r) \) involves polynomial expressions that reflect the non-uniformity of the charge distribution. The form of \( V(r) \) directly affects the derived electric field, as it requires taking the gradient, leading to the function for \( \vec{E} \).

For \( r \geq a \), the potential quickly drops to zero, indicating that outside this radius, the effects of the charge distribution vanish. Understanding the potential function's behavior is critical in predicting the impact of spherical charge distributions in various physical contexts.

Charge Density Calculations

Charge density \( \rho(r) \) represents the amount of charge per unit volume at a given point. Calculating \( \rho(r) \) is crucial for analyzing how charge influences the electric field and potential across different regions.

For \( r \leq a \), using Gauss's Law, we can express \( \rho(r) \) in terms of the electric field by recognizing \( abla \cdot \vec{E} = \frac{\rho(r)}{\epsilon_0} \). This calculation necessitates connecting the charge density with the derived electric field, revealing the specific behavior and contribution of \( \rho(r) \) to the overall field distribution.

For \( r \geq a \), since both the electric field and potential are zero, it follows that \( \rho(r) = 0 \). Showing that no charge density exists in this region is consistent with our understanding of confined charge systems. Calculating \( \rho(r) \) not only confirms the physical insights but also validates our boundary conditions and overall problem-solving approach.