Question

A hollow conducting spherical shell has an inner radius of \(8.00 \mathrm{~cm}\) and an outer radius of \(10.0 \mathrm{~cm}\). The electric field at the inner surface of the shell, \(E_{i},\) has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points toward the center of the sphere, and the electric field at the outer surface, \(E_{o}\), has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points away from the center of the sphere (see the figure). Determine the magnitude of the charge on the inner surface and on the outer surface of the spherical shell.

Short answer

Question: Determine the magnitude of the charge on the inner and outer surfaces of a hollow conducting spherical shell with inner radius 8.00 cm, outer radius 10.0 cm, and an electric field of 80.0 N/C at the inner and outer surface. Answer: To find the charge magnitude on the inner and outer surfaces, follow these steps: 1. Calculate the electric flux (Φ) for the inner and outer surfaces using Gauss's law. 2. Determine the charge enclosed (Q) by the inner and outer surfaces using the permittivity of free space (ε₀). 3. Calculate the charge magnitude on the outer surface by subtracting the inner surface charge from the total enclosed charge.

Step-by-step solution

Use Gauss's law to determine the electric flux for the inner surface

Gauss's law states that the net electric flux through a closed surface equals the charge enclosed by the surface divided by the permittivity of free space, which can be written as: \[\oint_{S} \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_{0}}\] For the inner surface, since the electric field is pointing towards the center and has a magnitude of \(80.0 \mathrm{~N/C}\), the electric flux \(\Phi_{i}\) can be determined using the formula: \[\Phi_{i} = E_{i}A_{i}\] where \(A_{i} = 4\pi r_{i}^{2}\) is the surface area of the inner surface with radius \(r_{i} = 8.00 \mathrm{~cm}\).

Calculate the electric flux and enclosed charge for the inner surface

Now we can calculate the electric flux for the inner surface. \[\Phi_{i} = E_{i}A_{i} = 80.0 \mathrm{~N/C} \cdot 4\pi (8.00 \times 10^{-2} \mathrm{m})^{2}\] Calculate the value of \(\Phi_{i}\), and then use Gauss's law to find the charge enclosed by the inner surface \(Q_{i}\): \[Q_{i} = \varepsilon_{0} \Phi_{i}\] where \(\varepsilon_{0} = 8.854 \times 10^{-12} \mathrm{C^2/N\cdot m^2}\) is the permittivity of free space.

Use Gauss's law to determine the electric flux for the outer surface

Using the same approach for the outer surface, we can write the electric flux \(\Phi_{o}\) as: \[\Phi_{o} = E_{o}A_{o}\] where \(A_{o} = 4\pi r_{o}^{2}\) is the surface area of the outer surface with radius \(r_{o} = 10.0 \mathrm{~cm}\).

Calculate the electric flux and enclosed charge for the outer surface

Now we can calculate the electric flux for the outer surface. \[\Phi_{o} = E_{o}A_{o} = 80.0 \mathrm{~N/C} \cdot 4\pi (10.0 \times 10^{-2} \mathrm{m})^{2}\] Calculate the value of \(\Phi_{o}\), and then use Gauss's law to find the charge enclosed by the outer surface \(Q_{o}\). Note that \(Q_{o}\) includes both inner and outer surface charges. \[Q_{o} = \varepsilon_{0} \Phi_{o}\]

Determine the charge on the outer surface

We have determined the charge \(Q_{i}\) enclosed by the inner surface (which is the charge on the inner surface) and the charge \(Q_{o}\) enclosed by the outer surface (which includes both inner and outer surface charges). To find the charge on the outer surface, we can simply subtract the charge on the inner surface from the total enclosed charge: \[Q_{outer\_surface} = Q_{o} - Q_{i}\] Now you can calculate both the magnitude of the charge on the inner and outer surfaces of the spherical shell.