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Chapter 22: Problem 21

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 flux through a spherical surface just inside the inner surface of the sphere?

Short Answer

Expert verified
(a) 7.61 x 10^{-6} C/m^2. (b) 345,200 N/C. (c) -56,465 N路m^2/C.

Step by step solution

01

Understanding the problem

We have a hollow conducting sphere with a given surface charge density on the outside. There's a new charge introduced in the cavity of the sphere, and we need to find the new charge density on the outside, the electric field just outside the sphere, and the electric flux inside the cavity.
02

Determine the new charge density on the outer surface

Initially, the surface charge density on the outer surface is \(+6.37 \times 10^{-6} \mathrm{C/m}^2\). The charge inside the cavity is \(-0.500 \times 10^{-6} \mathrm{C}\). Since the sphere is conductive, the charges will rearrange such that the inner surface holds charge equal in magnitude but opposite in sign to the inner cavity's charge (-0.500 碌C or -0.500 x 10^{-6} C). Therefore, the net additional charge on the outer surface becomes +0.500 碌C. Calculate new total outer surface charge: \(\sigma_{new} = \sigma_{old} + \dfrac{q_{inner}}{A} = 6.37 \times 10^{-6} \mathrm{C/m}^2 + \dfrac{0.500 \times 10^{-6} \mathrm{C}}{4\pi (0.250 \mathrm{m})^2}\).
03

Calculate the Electric Field just outside the sphere

Use Gauss's Law to calculate the electric field just outside the sphere. The total charge contributing to the electric field just outside is the charge on the outer surface. Using Gauss's law: \(E \cdot 4\pi r^2 = \dfrac{Q_{outer}}{\varepsilon_0}\), where \(r = 0.250 \mathrm{m}\), and \(\varepsilon_0 = 8.854 \times 10^{-12} \mathrm{C^2/N \cdot m^2}\). Solve for \(E\).
04

Calculate the Electric Flux just inside the inner surface

The electric flux through a surface just inside the inner surface is determined by the charge within that surface. In this case, Gauss's law tells us that flux \(\Phi = \dfrac{q_{enclosed}}{\varepsilon_0}\), where \(q_{enclosed} = -0.500 \times 10^{-6} \mathrm{C}\). Calculate using \(\varepsilon_0 = 8.854 \times 10^{-12} \mathrm{C^2/N \cdot m^2}\).

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Key Concepts

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

Conducting Sphere
Conducting spheres are fascinating objects to study because of how they manage electrical charges. When dealing with a conducting sphere, its surfaces become very important. In our exercise, the sphere has two surfaces: an outer and an inner surface.
  • The outer surface initially possesses a uniform charge density, meaning charge is evenly distributed across the surface.
  • Conductors allow rearrangement of charges. When a negative charge is introduced within the cavity (inner surface), the charges redistribute.
  • The charge inside the cavity influences the charge on the inner surface, creating equal and opposite charge on the inner lining to counterbalance.
As a result, any surplus or deficit charge affects the outer surface, changing its charge density. By introducing a charge inside the cavity, the outer surface gained an additional charge, adjusting the charge density. So, the charge on the inner surface migrates to neutralize the internal charge, redistributing any excess charge to the outer surface.
Electric Field
The electric field around a conductor is an essential concept. It helps us understand how charges affect regions in space. When working with a conducting sphere, calculating the electric field involves Gauss's Law.
  • The electric field just outside a sphere depends on surface charge.
  • According to Gauss's Law, the electric field (\(E\)) is derived from the total charge on the outer surface.
  • The exterior field only depends on outer surface charge as fields cancel inside a conductor.
To find the field strength, we employ Gauss's Law. Imagine a Gaussian surface just outside the sphere. This surface encloses the total outer charge, which defines the electric field strength at that point.
Gauss's Law states that the electric field is proportional to total enclosed charge. Thus, by solving the formula, one can achieve the electric field magnitude: \[E = \frac{Q_{outer}}{4\pi \varepsilon_0 r^2}\].
Electric Flux
Electric flux through a surface tells us how much electric field penetrates it. In our scenario, the flux is significant just inside the inner surface of the sphere. This is because the flux specifically considers the charge within enclosing surfaces.
  • Electric flux (\(\Phi\)) represents the amount of electric field passing through.
  • Gauss's Law gives that \(\Phi = \frac{q_{enclosed}}{\varepsilon_0}\).
  • For our hollow sphere, only the internal cavity charge contributes.
In essence, the electric flux relates directly to the internal charge. This flux does not depend on the actual spherical surface area, but rather the charge inside it. As long as the charge remains in the cavity, the internal flux stays constant, based on the internal charge through the formula \[\Phi = \frac{-0.500 \times 10^{-6} \mathrm{C}}{8.854 \times 10^{-12} \mathrm{C^2/N \cdot m^2}}\].

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Most popular questions from this chapter

A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho(r)\) given as follows: $$\begin{array}{ll}{\rho(r)=\rho_{0}(1-r / R)} & {\text { for } r \leq R} \\ {\rho(r)=0} & {\text { for } r \geq R}\end{array}$$ where \(\rho_{0}=3 Q / \pi R^{3}\) is a positive constant. (a) Show that the total charge contained in the charge distribution is \(Q .\) (b) Show that the electric field in the region \(r \geqslant R\) is identical to that produced by a point charge \(Q\) at \(r=0 .\) (c) Obtain an expression for the electric field in the region \(r \leq R .\) (d) Graph the electric-field magnitude \(E\) as a function of \(r .\) (e) Find the value of \(r\) at which the electric field is maximum, and find the value of that maximum field.

A negative charge \(-Q\) is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not? (c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge \(-Q ?\) Is it reasonable to say that the grounded conductor has shielded the region from the effects of the charge \(-Q ?\) In principle, could the same thing be done for gravity? Why or why not?

A long line carrying a uniform linear charge density \(+50.0 \mu C / m\) runs parallel to and 10.0 cm from the surface of a large, flat plastic sheet that has a uniform surface charge density of \(-100 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. Find the location of all points where an \(\alpha\) particle would feel no force due to this arrangement of charged objects.

A solid conducting sphere with radius \(R\) that carries positive charge \(Q\) is concentric with a very thin insulating shell of radius 2\(R\) that also carries charge \(Q .\) The charge \(Q\) is distributed uniformly over the insulating shell. (a) Find the electric field (magnitude and direction in each of the regions \(0 < r < R, R < r < 2 R,\) and \(r > 2 R .(\) b) Graph the electric-field magnitude as a function of \(r .\)

A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{m} .\) What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) \(0.500 \mathrm{m},(\mathrm{b}) 1.50 \mathrm{m},(\mathrm{c}) 2.50 \mathrm{m} ?\)
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