Charge $q$ is distributed uniformly throughout the volume of an insulating sphere of radius $R = 4.00$ cm. At a distance of $r = 8.00$ cm from the center of the sphere, the electric field due to the charge distribution has magnitude $E = 940$ N/C. What is the electric field at a distance of $2.00$ cm from the sphere's center?

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Understand that the problem involves a uniformly charged insulating sphere, which means we can use Gauss's Law to find the electric field at different distances from the center.

Recall Gauss's Law: $ \Phi_E = \frac{Q_{enc}}{\varepsilon_0} $, where $ \Phi_E $ is the electric flux, $ Q_{enc} $ is the enclosed charge, and $ \varepsilon_0 $ is the permittivity of free space.

For a point inside the sphere (at 2.00 cm from the center), the enclosed charge $ Q_{enc} $ is proportional to the volume of the sphere up to that radius. Use the formula $ Q_{enc} = \frac{q}{V_{total}} \times V_{enc} $, where $ V_{total} $ is the total volume of the sphere and $ V_{enc} $ is the volume enclosed by the radius 2.00 cm.

Calculate the volume of the sphere using $ V = \frac{4}{3} \pi R^3 $ and the volume enclosed by the radius 2.00 cm using $ V_{enc} = \frac{4}{3} \pi r^3 $.

Apply Gauss's Law to find the electric field $ E $ at 2.00 cm: $ E = \frac{Q_{enc}}{4\pi \varepsilon_0 r^2} $, where $ r $ is the distance from the center (2.00 cm in this case).

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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 charge within the surface, and ε₀ is the permittivity of free space. This law is crucial for calculating electric fields in symmetric charge distributions, such as spheres.

Electric Field Inside a Uniformly Charged Sphere

For a uniformly charged insulating sphere, the electric field inside the sphere (at a distance r from the center) is given by E = (1/4πε₀) * (Q_enclosed/r²), where Q_enclosed is the charge within a sphere of radius r. This field increases linearly with distance from the center until reaching the sphere's surface.

Charge Density and Enclosed Charge

Charge density (ρ) is the charge per unit volume, given by ρ = Q/V for a sphere. To find the charge enclosed within a smaller sphere of radius r, use Q_enclosed = ρ * (4/3)πr³. This concept helps determine the electric field at any point inside the sphere by calculating the charge enclosed within that radius.

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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

A conductor with an inner cavity, like that shown in Fig. $22.23$ c, carries a total charge of $positive 5.00$ nC. The charge within the cavity, insulated from the conductor, is $negative 6.00$ nC. How much charge is on (a) the inner surface of the conductor and (b) the outer surface of the conductor?

Textbook Question

An infinitely long cylindrical conductor has radius $r$ and uniform surface charge density $σ$ . In terms of $σ$ , what is the magnitude of the electric field produced by the charged cylinder at a distance $r > R$ from its axis? Then, express the result in terms of $λ$ and show that the electric field outside the cylinder is the same as if all the charge were on the axis.

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

An infinitely long cylindrical conductor has radius $r$ and uniform surface charge density $σ$ . In terms of $σ$ and $R$ , what is the charge per unit length $λ$ for the cylinder?

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