Question

Spherical surfaces at \(r=2,4\), and \(6 \mathrm{~m}\) carry uniform surface charge densities of \(20 \mathrm{nC} / \mathrm{m}^{2},-4 \mathrm{n} \mathrm{C} / \mathrm{m}^{2}\), and \(\rho_{\mathrm{so}}\), respectively. \((a)\) Find \(\mathbf{D}\) at \(r=1\), 3 , and \(5 \mathrm{~m}\). (b) Determine \(\rho_{S 0}\) such that \(\mathbf{D}=0\) at \(r=7 \mathrm{~m}\).

Short answer

In summary, we found the electric displacement field (D) at r = 1m, r = 3m, and r = 5m based on the Gauss's law for each sphere. The values of D are as follows: a) At r = 1m: \(\mathbf{D} = \frac{5}{ε_0} \cdot (20 \times 10^{-9}) \mathbf{\hat{r}}\) b) At r = 3m: \(\mathbf{D} = \frac{5}{ε_0} \cdot (16 \times 10^{-9}) \mathbf{\hat{r}}\) c) At r = 5m: \(\mathbf{D} = \frac{5}{ε_0} \cdot (16 \times 10^{-9}) \mathbf{\hat{r}}\) We also found the surface charge density ρ_s0 for the third sphere (r = 6m) so that the electric displacement field is zero at r = 7m: \(ρ_{s0} = -\frac{(20 \times 10^{-9})(4\pi(2)^2) + (-4 \times 10^{-9})(4\pi(4)^2)}{4\pi(6)^2}\)

Step-by-step solution

Find Electric Displacement Field for each sphere at the given distances

To find the electric displacement field D, we will use the Gauss's law for electric fields. The Gauss's law states that the electric flux through a closed surface is equal to the volume charge density divided by the permittivity of free space (\(ε_0\)). Gauss's Law: \(\oint \mathbf{D} \cdot d\mathbf{A} = \frac{Q}{ε_0}\) Now, we will find the electric displacement field (D) at r = 1m, r = 3m, and r = 5m using Gauss's law for each sphere. a) At r = 1m: This point is inside the first sphere (r = 2m) where charge density is 20nC/m². The sphere at r = 4m and r = 6m will not contribute to the electric field inside the first sphere. Therefore, we have: \(\oint \mathbf{D} \cdot d\mathbf{A} = \frac{20 \times 10^{-9}}{ε_0} \cdot (4 \pi (1)^2)\) Solving for D: \(\mathbf{D} = \frac{5}{ε_0} \cdot (20 \times 10^{-9}) \mathbf{\hat{r}}\) b) At r = 3m: This point is between the first and second spheres. The first sphere's charge density will still contribute to the electric displacement field, while the second sphere will contribute negatively. Therefore, we have: \(\oint \mathbf{D} \cdot d\mathbf{A} = \frac{(20 \times 10^{-9} - 4 \times 10^{-9})}{ε_0} \cdot (4 \pi (3)^2)\) Solving for D: \(\mathbf{D} = \frac{5}{ε_0} \cdot (16 \times 10^{-9}) \mathbf{\hat{r}}\) c) At r = 5m: This point is between the second and third sphere. The first two spheres' charge densities will contribute to the electric displacement field, while the third sphere will not contribute at this point. Therefore, we have: \(\oint \mathbf{D} \cdot d\mathbf{A} = \frac{(20 \times 10^{-9} - 4 \times 10^{-9})}{ε_0} \cdot (4 \pi (5)^2)\) Solving for D: \(\mathbf{D} = \frac{5}{ε_0} \cdot (16 \times 10^{-9}) \mathbf{\hat{r}}\)

Determine the surface charge density (ρ_s0) such that D = 0 at r = 7m

We want the electric displacement field to be zero at r = 7m, which means the contributions from all spheres at this point should be cancelled. We need to find the surface charge density ρ_s0 for the third sphere (r = 6m) such that D = 0 at r = 7m. As the total D = 0, we can write the equation: \(\frac{Q_1 + Q_2 + Q_3}{ε_0 \cdot 4\pi(7)^2} = 0\) Plug in the charge densities and radii for the first two spheres (Q = charge density * surface area): \(\frac{(20 \times 10^{-9})(4\pi(2)^2) + (-4 \times 10^{-9})(4\pi(4)^2) + ρ_{s0}(4\pi(6)^2)}{ε_0 \cdot 4\pi(7)^2} = 0\) Solve for ρ_s0: \(ρ_{s0} = -\frac{(20 \times 10^{-9})(4\pi(2)^2) + (-4 \times 10^{-9})(4\pi(4)^2)}{4\pi(6)^2}\) Therefore, the surface charge density ρ_s0 for the third sphere (r = 6m) that makes the electric displacement field equal to zero at r = 7m is found.

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism. It connects the electric field surrounding a charged object to the charge enclosed within a given surface. According to Gauss's Law, the electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε_0). Mathematically, this is expressed as:

• \(\oint \mathbf{D} \cdot d\mathbf{A} = \frac{Q}{ε_0}\)

The concept of electric flux is crucial here. Imagine a surface surrounding a charged object. The flux represents the number of electric field lines crossing that surface. More field lines crossing mean more charge is enclosed.

In practical terms, Gauss's Law allows us to calculate electric fields for symmetrical charge distributions. By applying this law, we determine the electric displacement field (D), often simplifying complex problems when symmetry is present, such as with spherical charge distributions.

Surface Charge Density

Surface charge density, denoted by \(\sigma\), describes how charge is distributed over a surface. It's defined as the charge per unit area, expressed in units like nanocoulombs per square meter (nC/m²).

• Higher surface charge density means more charge is packed into a smaller surface area.
• This can lead to stronger electric fields emanating from the surface.

In the given exercise, surface charge densities help us understand how charges are behaving on different spherical surfaces. By knowing these densities, we can apply Gauss's Law to determine the resulting electric field or displacement field in and around these surfaces.

Thinking about surface charge density provides insight into how charges interact with their surroundings, affecting the overall electric field's strength and direction. This is particularly important when charges vary across multiple surfaces.

Permittivity of Free Space

Permittivity of free space, symbolized as \(ε_0\), is a natural constant that measures the ability of the vacuum to permit electric field lines. This constant is crucial when discussing electric fields and forces in the vacuum.

• It has a value of approximately \(8.854 \times 10^{-12} \text{ F/m}\) (farads per meter).
• It appears in Coulomb's Law, Gauss's Law, and equations for capacitance.

The role of \(ε_0\) in Gauss’s Law is vital. It acts as a scaling factor for electric fields and is used when determining the displacement field \(\mathbf{D}\). The permittivity of free space ensures the units are consistent and aligns the field's strength with the actual physical charge.

Overall, \(ε_0\) is a cornerstone value in electromagnetics, allowing physicists and engineers to predict how charges will act in non-material environments.

Spherical Surfaces

Spherical surfaces are a common structure in electromagnetic problems, often simplifying calculations due to their symmetrical nature. Imagine concentric circles in three dimensions, where charge is distributed evenly.

• In the context of Gauss's Law, spherical symmetry means electric fields only depend on the radius \(r\), making it easier to apply the law.
• Such symmetry leads to useful simplifications, allowing us to "ignore" angles when integrating over a surface.

Applying Gauss's Law to spherical surfaces involves considering these surfaces' areas. As in the original exercise, spheres at radii of 2m, 4m, and 6m hold different charge densities. This requires calculating the effect each has on the displacement field at various radii.

Spherical surfaces exemplify ideal scenarios where theoretical models and real-world problems meet; they allow clarity in calculations and in understanding how charges interact at different distances.