Question

(A) Calculate the magnitude of the electric field at a point outside the sphere. Gaussian sphere Conceptualize: Note how this problem differs from our previous discussion of Gauss's law. Now we are considering the electric field due to a distribution of charge. We found the field for various distributions of charge in the chapter entitled Electric Fields by integrating over the distribution. In this chapter, we find the electric field using Gauss's law. A uniformly charged insulating sphere of radius a and total charge Q. (a) For points outside the sphere, a large, spherical gaussian surface is drawn concentric with the sphere. In diagrams such as this one, the dotted line represents the intersection of the gaussian surface with the plane of the page. (b) For points inside the sphere, a spherical gaussian surface smaller than the sphere is drawn. Categorize: Because the charge is distributed uniformly throughout the sphere, the charge distribution has spherical symmetry and we can apply Gauss's law to find the electric field. Analyze: To reflect the spherical symmetry, let's choose a spherical gaussian surface of radius r, concentric with the sphere, as shown in Figure (a). For this choice, E and dA are parallel everywhere on the surface and E · dA = EdA. Replace E · dA in Gauss's Law with EdA: ?_E = ? E · dA = ? EdA = Q/?_0 By symmetry, E is constant everywhere on the surface, so we can remove E from the integral: ? EdA = E ? dA = E(4?r^2) = Q/?_0 Solve for E. (1) E = Q/(4??_0r^2) (Use the following as necessary: k_e, Q, and r.) E = k_e Q / r^2 (for r > a)

          (A) Calculate the magnitude of
the electric field at a point
outside the sphere.

Gaussian
sphere

Conceptualize: Note how this
problem differs from our
previous discussion of Gauss's
law. Now we are considering
the electric field due to a
distribution of charge. We
found the field for various distributions of charge in the chapter entitled Electric Fields by integrating over the distribution. In this chapter, we find the electric field using Gauss's law.

A uniformly charged insulating sphere of radius a and total charge Q. (a) For points outside the sphere, a large, spherical gaussian surface is drawn concentric with the sphere. In diagrams such as this one, the dotted line represents the intersection of the gaussian surface with the plane of the page. (b) For points inside the sphere, a spherical gaussian surface smaller than the sphere is drawn.

Categorize: Because the charge is distributed uniformly throughout the sphere, the charge distribution has spherical symmetry and we can apply Gauss's law to find the electric field.

Analyze: To reflect the spherical symmetry, let's choose a spherical gaussian surface of radius r,
concentric with the sphere, as shown in Figure (a). For this choice, E and dA are parallel everywhere on
the surface and E &#183; dA = EdA.

Replace E &#183; dA in Gauss's Law with EdA:
?_E = ? E &#183; dA = ? EdA = Q/?_0

By symmetry, E is constant everywhere on
the surface, so we can remove E from the
integral:
? EdA = E ? dA = E(4?r^2) = Q/?_0

Solve for E.
(1) E = Q/(4??_0r^2)
(Use the following as necessary: k_e, Q, and
r.)

E = k_e Q / r^2 (for r > a)

(A) Calculate the magnitude of
the electric field at a point
outside the sphere.

Gaussian
sphere

Conceptualize: Note how this
problem differs from our
previous discussion of Gauss's
law. Now we are considering
the electric field due to a
distribution of charge. We
found the field for various distributions of charge in the chapter entitled Electric Fields by integrating over the distribution. In this chapter, we find the electric field using Gauss's law.

A uniformly charged insulating sphere of radius a and total charge Q. (a) For points outside the sphere, a large, spherical gaussian surface is drawn concentric with the sphere. In diagrams such as this one, the dotted line represents the intersection of the gaussian surface with the plane of the page. (b) For points inside the sphere, a spherical gaussian surface smaller than the sphere is drawn.

Categorize: Because the charge is distributed uniformly throughout the sphere, the charge distribution has spherical symmetry and we can apply Gauss's law to find the electric field.

Analyze: To reflect the spherical symmetry, let's choose a spherical gaussian surface of radius r,
concentric with the sphere, as shown in Figure (a). For this choice, E and dA are parallel everywhere on
the surface and E #183; dA = EdA.

Replace E #183; dA in Gauss's Law with EdA:
?E = ? E #183; dA = ? EdA = Q/?0

By symmetry, E is constant everywhere on
the surface, so we can remove E from the
integral:
? EdA = E ? dA = E(4?r^2) = Q/?0

Solve for E.
(1) E = Q/(4??0r^2)
(Use the following as necessary: ke, Q, and
r.)

E = ke Q / r^2 (for r > a)

Added by Melissa W.

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