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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?

Short Answer

Expert verified
(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 by step solution

01

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: \( Q_{\text{outer initial}} = 6.37 \times 10^{-6} \, \mathrm{C/m}^2 \times 4 \pi (0.250 \, \mathrm{m})^2 = 5.00 \times 10^{-6} \, \mathrm{C} \).
02

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 \( Q_{\text{outer}} = Q_{\text{outer initial}} + (-Q_{\text{cavity}}) = 5.00 \times 10^{-6} \, \mathrm{C} + 0.500 \times 10^{-6} \, \mathrm{C} = 5.50 \times 10^{-6} \, \mathrm{C} \).
03

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:\( \sigma_{\text{new}} = \frac{5.50 \times 10^{-6} \, \mathrm{C}}{4 \pi (0.250 \, \mathrm{m})^2} = 7.01 \times 10^{-6} \, \mathrm{C/m}^2 \).
04

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:\( E = \frac{7.01 \times 10^{-6} \, \mathrm{C/m}^2}{8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2}} = 7.92 \times 10^5 \, \mathrm{N/C} \).
05

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:\( \Phi = \frac{-0.500 \times 10^{-6} \, \mathrm{C}}{8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2}} = -5.65 \times 10^4 \, \mathrm{N \cdot m^2/C} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

22\. 14. Electric Fields in an Atom. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{m}\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{m} ?(\mathrm{c})\) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

22.56. Can Electric Forces Alone Give Stable Rquilibrium? In Chapter \(21,\) several examples were given of calculating the force exerted on a point charge by other point charges in its sur- roundings. (a) Consider a positive point charge \(+q .\) Give an example of how you would place two other point charges of your choosing so that the net force on charge \(+q\) will be zero. \((b)\) If the net force on charge \(+q\) is zero, then that charge is in equilibrium. The equilibrium will be stable if, when the charge \(+q\) is displaced slightly in any direction from its position of equilibrium, the net force on the charge pushes it back toward the equilibrium position. For this to be the case, what must the direction of the electric field \(\overrightarrow{\boldsymbol{E}}\) be due to the other charges at points surrounding the equilibrium position of \(+q ?(c)\) Imagine that the charge \(+q\) is moved very far away, and imagine a small Gaussian surface centered on the position where \(+q\) was in equilibrium. By applying Gauss's law to this surface, show that it is impossible to satisfy the condition for stability described in part (b). In other words, a charge \(+q\) cannot be held in stable equilibrium by electrostatic forces alone. This result is known as Earnshaw's theorem. (d) Parts (a)-(c) referred to the equilibrium of a positive point charge \(+q\) . Prove that Earnshaw's theorem also applies to a negative point charge \(-q\) .

22.5. A hemispherical surface with radius \(r\) in a region of uniform electric field \(\vec{E}\) has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

22.9. A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter \(12.0 \mathrm{cm},\) giving it a charge of \(-15.0 \mu \mathrm{C}\) . Find the electric field (a) just inside the paint layer; (b) just outside the paint layer: (c) 5.00 \(\mathrm{cm}\) outside the surface of the paint layer.

22.3 You measure an electric field of \(1.25 \times 10^{6} \mathrm{N} / \mathrm{C}\) at a distance of 0.150 \(\mathrm{m}\) from a point charge. (a) What is the electric flux through a sphere at that distance from the charge? (b) What is the magnitude of the charge?

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