Problem 23 - University Physics with Modern Physics

Question

22.23. A hollow, conducting sphere with an outer radius of 0.250 \(\mathrm{m}\) and an inner radius of 0.200 \(\mathrm{m}\) has a uniform surface charge density of \(+6.37 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) A charge of \(-0.500 \mu \mathrm{C}\) is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric fiux through a spherical surface just inside the inner surface of the sphere?

Step-by-Step Solution

Answer

(a) The new charge density is \( 7.01 \times 10^{-6} \, \mathrm{C/m^2} \). (b) The electric field just outside is \( 7.92 \times 10^5 \, \mathrm{N/C} \). (c) The electric flux through the inner surface is \(-5.65 \times 10^4 \, \mathrm{N \cdot m^2/C} \).

Step 1: Determine the Total Charge on the Outside Surface of the Sphere

The initial charge on the outer surface is given by the charge density \( \sigma = +6.37 \times 10^{-6} \mathrm{C/m}^2 \). To find the total charge on the outer surface, multiply the charge density by the outer surface area. The outer area is \( 4 \pi (R_{\text{outer}})^2 = 4 \pi (0.250 \, \mathrm{m})^2 \). Thus, the charge is \( Q_{\text{outer initial}} = \sigma \cdot 4\pi (0.250 \, \mathrm{m})^2 \).Calculate: \( Q_{\text{outer initial}} = 6.37 \times 10^{-6} \, \mathrm{C/m}^2 \times 4 \pi (0.250 \, \mathrm{m})^2 = 5.00 \times 10^{-6} \, \mathrm{C} \).

Step 2: Account for the Introduced Charge in the Inner Cavity

When a \( -0.500 \mu \mathrm{C} \) charge is introduced into the cavity, the inner surface will have a charge of \( +0.500 \mu \mathrm{C} \) due to induction (to neutralize the inner cavity), and the outer surface must adjust to maintain overall neutrality of the object. Therefore, the new total charge on the outer surface becomes \( Q_{\text{outer}} = Q_{\text{outer initial}} + (-Q_{\text{cavity}}) = 5.00 \times 10^{-6} \, \mathrm{C} + 0.500 \times 10^{-6} \, \mathrm{C} = 5.50 \times 10^{-6} \, \mathrm{C} \).

Step 3: Calculate the New Charge Density on the Outside of the Sphere

The new charge density on the outer surface can be calculated using the updated total charge and the outer surface area:\( \sigma_{\text{new}} = \frac{Q_{\text{outer}}}{4 \pi (R_{\text{outer}})^2} \).Substitute the values:\( \sigma_{\text{new}} = \frac{5.50 \times 10^{-6} \, \mathrm{C}}{4 \pi (0.250 \, \mathrm{m})^2} = 7.01 \times 10^{-6} \, \mathrm{C/m}^2 \).

Step 4: Determine the Electric Field Just Outside the Sphere

The electric field just outside the conducting sphere is given by\( E = \frac{\sigma_{\text{new}}}{\varepsilon_0} \), where \( \varepsilon_0 \) is the permittivity of free space \( 8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2} \).Calculate:\( E = \frac{7.01 \times 10^{-6} \, \mathrm{C/m}^2}{8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2}} = 7.92 \times 10^5 \, \mathrm{N/C} \).

Step 5: Calculate the Electric Flux Through the Inner Surface

According to Gauss's law, the electric flux \( \Phi \) through a closed surface is given by \( \Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \). For a surface just inside the inner surface of the sphere, the enclosed charge is the introduced \( -0.500 \mu \mathrm{C} \).Calculate:\( \Phi = \frac{-0.500 \times 10^{-6} \, \mathrm{C}}{8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2}} = -5.65 \times 10^4 \, \mathrm{N \cdot m^2/C} \).

Key Concepts

Charge Density

Charge density is a measure of how much electric charge is accumulated on a surface or within a volume. It is typically expressed as a ratio of charge to area for surfaces (denoted as \(\sigma\)) or charge to volume for volumes (denoted as \(\rho\)). In this exercise, the charge density relates to the surface of a conducting sphere.

To find the initial charge on the outer surface of the sphere, we multiply the given charge density by the surface area of the sphere. The surface area of a sphere is calculated using the formula:

• \(A = 4 \pi R^2\)

where \(R\) is the radius of the sphere.

In the exercise, the outer charge density changes when a charge is added to the inner cavity. The outer surface must adjust to accommodate this change to maintain neutrality. The new charge density can be calculated by dividing the new total outer charge by the same surface area.
Understanding charge density is crucial because it affects the electric field generated around charged objects.

Electric Field

An electric field is a vector field around a charged object where a force would be exerted on other charges present. The strength of this field, \(E\), at a point outside a charged conductor, is related directly to the surface charge density \(\sigma\) and inversely to the permittivity of free space \(\varepsilon_0\).

The electric field just outside a conductor is given by:

• \(E = \frac{\sigma}{\varepsilon_0}\)

where \(\varepsilon_0\) is approximately equal to \(8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2}\).

This relation illustrates that the electric field is stronger with higher charge densities and weaker with a larger permittivity. This field is important in understanding how forces will act on charges in the vicinity of charged surfaces. Calculating the electric field is essential because it determines how other charges will move in the presence of the charged conductor.

Gauss's Law

Gauss's Law is a fundamental principle in electrostatics that links the electric flux passing through a closed surface to the charge enclosed by that surface. Mathematically, it is expressed as:

• \(\Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\)

where \(\Phi\) is the electric flux and \(Q_{\text{enclosed}}\) is the total charge within the closed surface.

In the given exercise, Gauss's Law is applied to a surface just inside the inner cavity of the conducting sphere. By applying Gauss’s Law, we can determine the electric flux based on the enclosed charge. Even if the actual shape of the charged body is complex, Gauss's Law helps in simplifying the calculations significantly, especially for symmetrical shapes like spheres or cylinders.

This law is a powerful tool in electrostatics allowing us to calculate electric fields and flux when direct measurement and observation are challenging.

Electric Flux

Electric flux is a concept used to measure the amount of electric field passing through a given surface. In the context of this problem, it helps us understand how the electric field behaves inside the cavity of a conducting sphere.

Electric flux \(\Phi\) through a surface can be calculated from Gauss's Law:

• \(\Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\)

where \(Q_{\text{enclosed}}\) is the charge within the considered surface.

In the exercise, even though the sphere’s inner surface encloses a negative charge, the key point is quantifying the electric field impact within the conducting cavity. This ensures students grasp how the presence of charge affects the field dynamics around it. Electric flux is critical in understanding field lines and their interaction with matter. It provides a broader perspective on how fields interact with various geometries beyond simply the force they exert.