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Suppose the charge at the center is now increased to +6Q, while the charge at the surface of the conductor is changed to +4Q. (Use the following as necessary: ?0, ?, Q and r.) (a) Find the electric field exterior to the sphere, for r > b.

Suppose the charge at the center is now increased to +6Q, while the charge at the surface of the conductor is changed to +4Q. (Use the following as necessary: ?0, ?, Q and r.) (a) Find the electric field exterior to the sphere, for r > b.

College Physics
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Suppose the charge at the center is now increased to +6Q, while the charge at the surface of the conductor is changed to +4Q. (Use the following as necessary: ?0, ?, Q and r.)

(a) Find the electric field exterior to the sphere, for r > b.
EXAMPLE 15.7 The Electric Field of a Charged Spherical Shell GOAL Use Gauss's law to determine electric fields when the symmetry is spherical. Consider the following figures. Gaussian Gaussian surface surface 王 a a 9. Ein = 0 a (a) The electric field inside a uniformly charged spherical shell is zero. It is also zero for the conducting material in the region a <r < b. The field outside is the same as that of a point charge having a total charge Q located at the center of the shell. (b) The construction of a Gaussian surface for calculating the electric field inside a spherical shell. (c) The construction of a Gaussian surface for calculating the electric field outside a spherical shell. PROBLEM A spherical conducting shell of inner radius a and outer radius b carries a total charge +Q distributed on the surface of a conducting shell (figure a). The quantity Q is taken to be positive. (a) Find the electric field in the interior of the conducting shell, for r < a, and (b) the electric field outside the shell, for r > b. (c) If an additional charge of -2Q is placed at the center, find the electric field for r > b. (d) What is the distribution of charge on the sphere in part (c)? STRATEGY For each part, draw a spherical Gaussian surface in the region of interest. Add up the charge inside the Gaussian surface, substitute it and the area into Gauss's law, and solve for the electric field. To find the distribution of charge in part (c), use Gauss's law in reverse: the charge distribution must be such that the electrostatic field is zero inside a conductor. SOLUTION (a) Find the electric field for r < a. Apply Gauss's law to the Gaussian Qinside EA = E(4xr²) = = 0 → E = 0 surface illustrated in figure b (note that there isn't any charge inside this surface).
Transcribed Image Text:EXAMPLE 15.7 The Electric Field of a Charged Spherical Shell GOAL Use Gauss's law to determine electric fields when the symmetry is spherical. Consider the following figures. Gaussian Gaussian surface surface 王 a a 9. Ein = 0 a (a) The electric field inside a uniformly charged spherical shell is zero. It is also zero for the conducting material in the region a <r < b. The field outside is the same as that of a point charge having a total charge Q located at the center of the shell. (b) The construction of a Gaussian surface for calculating the electric field inside a spherical shell. (c) The construction of a Gaussian surface for calculating the electric field outside a spherical shell. PROBLEM A spherical conducting shell of inner radius a and outer radius b carries a total charge +Q distributed on the surface of a conducting shell (figure a). The quantity Q is taken to be positive. (a) Find the electric field in the interior of the conducting shell, for r < a, and (b) the electric field outside the shell, for r > b. (c) If an additional charge of -2Q is placed at the center, find the electric field for r > b. (d) What is the distribution of charge on the sphere in part (c)? STRATEGY For each part, draw a spherical Gaussian surface in the region of interest. Add up the charge inside the Gaussian surface, substitute it and the area into Gauss's law, and solve for the electric field. To find the distribution of charge in part (c), use Gauss's law in reverse: the charge distribution must be such that the electrostatic field is zero inside a conductor. SOLUTION (a) Find the electric field for r < a. Apply Gauss's law to the Gaussian Qinside EA = E(4xr²) = = 0 → E = 0 surface illustrated in figure b (note that there isn't any charge inside this surface).
QUESTION If the charge at the center of the sphere is made positive, how is the charge on the inner surface of the sphere affected? (Select all that apply.) The absolute value of the charge on the inner surface would decrease. The charge on the inner surface would be positive. The charge on the inner surface would be negative. The absolute value of the charge on the inner surface would remain the same. There would be no net charge on the inner surface. The absolute value of the charge on the inner surface would increase. EXERCISE HINTS: GETTING STARTED I'M STUCK! Suppose the charge at the center is now increased to +6Q, while the charge at the surface of the conductor is changed to +4Q. (Use the following as necessary: E9, T, Q and r.) (a) Find the electric field exterior to the sphere, for r > b. 100 ,2 E = (b) What's the electric field inside the conductor, for a < r < b. E = (c) Find the charge distribution on the conductor. inner surface -6Q outer surface +10Q
Transcribed Image Text:QUESTION If the charge at the center of the sphere is made positive, how is the charge on the inner surface of the sphere affected? (Select all that apply.) The absolute value of the charge on the inner surface would decrease. The charge on the inner surface would be positive. The charge on the inner surface would be negative. The absolute value of the charge on the inner surface would remain the same. There would be no net charge on the inner surface. The absolute value of the charge on the inner surface would increase. EXERCISE HINTS: GETTING STARTED I'M STUCK! Suppose the charge at the center is now increased to +6Q, while the charge at the surface of the conductor is changed to +4Q. (Use the following as necessary: E9, T, Q and r.) (a) Find the electric field exterior to the sphere, for r > b. 100 ,2 E = (b) What's the electric field inside the conductor, for a < r < b. E = (c) Find the charge distribution on the conductor. inner surface -6Q outer surface +10Q
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