Physics Solution Manual for 1100 and 2101

2 as the force f that acts on a test charge q0

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Unformatted text preview: the distance from the common center of the shells and will be the distance at which we are to evaluate the electric field. Because the electric field has spherical symmetry, the magnitude E of the electric field is constant at all points on any such spherical Gaussian surface. Furthermore, the electric field is directed either radially outward (if the net charge within the Gaussian surface is positive) or radially inward (if the net charge within the Gaussian surface is negative). This means that the angle φ between the electric field and the normal to any such spherical Gaussian surface is either 0.0° or 180°. The quantity E cos φ , therefore, is constant, and may be factored out of the summation in Equation 18.6: Φ E = Σ ( E cos φ ) ∆A = ( E cos φ ) Σ∆A (1) The sum Σ∆A of all the tiny sections of area ∆A that compose a spherical Gaussian surface is the total surface area Σ∆A = A = 4π r 2 of a sphere of radius r. Thus, Equation (1) becomes Φ E = ( E cos φ ) Σ∆A = 4π r 2 E cos φ (2) Chapter 18 Problems 993 SOLUTION a. The outer shell has a radius r2 = 0.15 m, and we are to determine the electric field at a distance r = 0.20 m from the common center of the shells. Therefore, we will choose a spherical Gaussian surface (radius r = 0.20 m) that encloses both shells and shares their common center. According to Gauss’ law and Equation (2), we have that the net electric flux ΦE through this sphere is Σ ( E cos φ ) ∆A = 4π r 2 E cos φ = Q (3) ε0 Solving Equation (3) for E yields E= Q (4) 4πε 0 r 2 cos φ Because the chosen Gaussian surface encloses both shells, the net charge Q enclosed by the surface is Q = q1 + q2. The positive charge q2 on the outer shell has a larger magnitude than the negative charge q1 on the inner shell, so that Q is a positive net charge. Therefore, the electric field is directed radially outward , and the angle between the electric field and the normal to the surface of the spherical Gaussian surface is φ = 0.0°. Therefore, Equation (4) gives the electric field magnitude as E= q1 + q2 Q −1.6 × 10 −6 C + 5.1× 10 −6 C = = 4πε 0 r 2 cos φ 4πε 0 r 2 cos φ 4π 8.85 × 10 −12 C 2 / N ⋅ m 2 ( 0.20 m )2 cos 0.0o ( ) = 7.9 × 105 N/C b. We again choose a spherical Gaussian surface concentric with the shells, this time of radius r = 0.10 m. The radius of this sphere is greater than the radius (r1 = 0.050 m) of the inner shell but less than the radius (r2 = 0.15 m) of the outer shell. Therefore, this Gaussian surface is located between the two shells and encloses only the charge on the inner shell: Q = q1. This is a negative charge, so that the electric field is directed radially inward , and the angle between the electric field and the normal to the surface of the Gaussian sphere is φ = 180°. From Equation (4), then, we have that E= q1 Q −1.6 × 10 −6 C = = 4πε 0 r 2 cos φ 4πε 0 r 2 cos φ 4π 8.85 × 10 −12 C 2 / N ⋅ m 2 ( 0.10 m )2 cos180o = 1.4 × 106 N/C ( ) 994 ELECTRIC FORCES AND ELECTRIC FIELDS c. Choosing a spherical Gaussian surface with a radius of r = 0.025 m, we see that it is entirely inside the inner shell (r1 = 0.050 m). Therefore, the enclosed charge is zero: Q = 0 C. Equation (4) shows that the electric field at this distance from the common center is zero: E= Q 0C = = 0 N/C 2 4πε 0 r cos φ 4πε 0 r 2 cos φ 61. REASONING We use a Gaussian surface that is a sphere centered within the solid sphere. The radius r of this surface is smaller than the radius R of the solid sphere. Equation 18.7 gives Gauss’ law as follows: Q Σ ( E cos φ ) ∆ A = (18.7) 14 244 ε 0 4 3 Electric flux, Φ E We will deal first with the left side of this equation and evaluate the electric flux ФE. Then we will evaluate the net charge Q within the Gaussian surface. SOLUTION The positive charge is Normal spread uniformly throughout the solid sphere and, therefore, is spherically Angle φ between E E symmetric. Consequently, the electric and the normal is 0º field is directed radially outward, and for each element of area ∆ A is perpendicular to the surface. This means that the angle φ between the normal to the surface and the field is 0º, as the drawing shows. Furthermore, the electric field has the same magnitude everywhere on the Gaussian surface. Because of these considerations, we can write the electric flux as follows: Σ ( E cos φ ) ∆ A = Σ ( E cos 0° ) ∆ A = E ( Σ∆ A ) The term Σ∆ A is the entire area of the spherical Gaussian surface or 4πr2. With this substitution, the electric flux becomes ( Σ ( E cos φ ) ∆ A = E ( Σ∆ A) = E 4π r 2 ) (1) Now consider the net charge Q within the Gaussian surface. This charge is the charge density times the volume 4 π r 3 encompassed by that surface: 3 Q= ( ) q qr 3 × 4 π r3 = 3 3 4 π R3 R 32 44 1 3 123 Charge density Volume of Gaussian surface (2) Chapter 18 Problems 995 Substituting Equations (1) and (2) into Equation 18.7 gives ( ) E 4π r 2 = qr 3 / R3...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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