A hollow, conducting sphere with an outer radius of 0.2500.250 m and an inner radius of 0.2000.200 m has a uniform surface charge density of +6.37×10−6+6.37\$$times10^{-6} C/m2. A $$charge of −0.500−0.500 μ\muC is now introduced at the center of the cavity inside the sphere. What is the new charge density on the outside of the sphere?

Ch 22: Gauss' Law

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

All textbooks

Young & Freedman Calc 14th Edition

Ch 22: Gauss' Law

Problem 19a

Chapter 22, Problem 19a

A hollow, conducting sphere with an outer radius of $0.250$ m and an inner radius of $0.200$ m has a uniform surface charge density of $+ 6.37 \times 10^{- 6}$ C/m². A charge of $negative 0.500$ $μ$ C is now introduced at the center of the cavity inside the sphere. What is the new charge density on the outside of the sphere?

Verified step by step guidance

First, understand that the hollow conducting sphere will redistribute its charges to maintain electrostatic equilibrium. The charge inside the cavity will induce an equal and opposite charge on the inner surface of the sphere.

Calculate the total charge induced on the inner surface of the sphere. Since the charge inside the cavity is -0.500 μC, the inner surface will have a charge of +0.500 μC to neutralize the electric field inside the conductor.

Determine the total charge on the outer surface of the sphere. The original charge on the outer surface is determined by the surface charge density and the surface area of the outer sphere. Use the formula: $ Q_{outer} = \sigma \times 4\pi R_{outer}^2 $, where $ \sigma = 6.37 \times 10^{-6} \text{ C/m}^2 $ and $ R_{outer} = 0.250 \text{ m} $.

Add the induced charge from the inner surface to the original charge on the outer surface to find the new total charge on the outer surface. The total charge on the outer surface is the sum of the original charge and the charge induced by the inner cavity charge.

Finally, calculate the new surface charge density on the outer surface using the formula: $ \sigma_{new} = \frac{Q_{outer, new}}{4\pi R_{outer}^2} $, where $ Q_{outer, new} $ is the total charge calculated in the previous step.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It is expressed as Φ = Q_enclosed/ε₀, where Φ is the electric flux, Q_enclosed is the total charge within the surface, and ε₀ is the permittivity of free space. This law is crucial for understanding how charges distribute on conductors and how they affect electric fields.

Conductors in Electrostatic Equilibrium

In electrostatic equilibrium, the electric field inside a conductor is zero, and any excess charge resides on the surface. This is because charges within a conductor will move until they reach a state where the internal electric field cancels out. This principle helps determine how charges distribute on the inner and outer surfaces of a hollow conductor.

Surface Charge Density

Surface charge density, denoted by σ, is the amount of charge per unit area on a surface. It is calculated as σ = Q/A, where Q is the total charge and A is the area. For a spherical conductor, the surface charge density can change if additional charges are introduced, affecting the distribution of charge on the outer surface.

Recommended video:

Guided course

04:03

Surface Charge Density

Related Practice

Textbook Question

Charge $q$ is distributed uniformly throughout the volume of an insulating sphere of radius $R equals 4.00$ cm. At a distance of $r equals 8.00$ cm from the center of the sphere, the electric field due to the charge distribution has magnitude $E equals 940$ N/C. What is the volume charge density for the sphere?

Textbook Question

A hollow, conducting sphere with an outer radius of $0.250$ m and an inner radius of $0.200$ m has a uniform surface charge density of $+6.37$\times$10^{-6}$ C/m². A charge of $−0.500$ $\mu$ C is now introduced at the center of the cavity inside the sphere. What is the electric flux through a spherical surface just inside the inner surface of the sphere?

Textbook Question

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of $-3.63$\times$10^{16}$ Nm²/C at the planet's surface. Calculate the electric field at the planet's surface (refer to the astronomical data inside the back cover).

views

Textbook Question

A very long uniform line of charge has charge per unit length $4.80$ $μ$ C/m and lies along the $x$ -axis. A second long uniform line of charge has charge per unit length $negative 2.40$ $μ$ C/m and is parallel to the $x$ -axis at $y equals 0.400$ m. What is the net electric field (magnitude and direction) at the following points on the $y$ -axis: (a) $y equals 0.200$ m and (b) $y equals 0.600$ m?

views

Textbook Question

A hollow, conducting sphere with an outer radius of $0.250$ m and an inner radius of $0.200$ m has a uniform surface charge density of $+6.37$\times$10^{-6}$ C/m². A charge of $−0.500$ $\mu$ C is now introduced at the center of the cavity inside the sphere. Calculate the strength of the electric field just outside the sphere?

Textbook Question

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of $-3.63$\times$10^{16}$ Nm²/C at the planet's surface. Calculate the charge density on Mars, assuming all the charge is uniformly distributed over the planet's surface.

views