Chapter 22 HW Solutions

# Chapter 22 HW Solutions - PHYS 241 Ch 22 Homework Solutions...

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Unformatted text preview: PHYS 241 Ch. 22 Homework Solutions 22.7.(a) I DENTIFY : Use Eq.(22.5) to calculate the flux through the surface of the cylinder. S ET U P : The line of charge and the cylinder are sketched in Figure 22.7. Figure 22.7 E XECUTE : The area of the curved part of the cylinder is 2 . A rl π = The electric field is parallel to the end caps of the cylinder, so ⋅ = E A r r for the ends and the flux through the cylinder end caps is zero. The electric field is normal to the curved surface of the cylinder and has the same magnitude /2 E r λ π = P at all points on this surface. Thus φ = ° and ( 29 ( 29 ( 29 ( 29 6 5 2 12 2 2 6.00 10 C/m 0.400 m cos /2 2 2.71 10 N m /C 8.854 10 C / N m E l EA EA r rl λ φ λ π π-- × Φ = = = = = = × ⋅ × ⋅ P P (b) In the calculation in part (a) the radius r of the cylinder divided out, so the flux remains the same, 5 2 2.71 10 N m /C. E Φ = × ⋅ (c) ( 29 ( 29 6 5 2 12 2 2 6.00 10 C/m 0.800 m 5.42 10 N m /C 8.854 10 C / N m E l λ-- × Φ = = = × ⋅ × ⋅ P (twice the flux calculated in parts (b) and (c)). E VALUATE : The flux depends on the number of field lines that pass through the surface of the cylinder. 22.9.I DENTIFY : Apply the results in Example 21.10 for the field of a spherical shell of charge. S ET U P : Example 22.10 shows that E = inside a uniform spherical shell and that 2 q E k r = outside the shell. E XECUTE : (a) E = (b) 0.060 m r = and 6 9 2 2 7 2 15.0 10 C (8.99 10 N m /C ) 3.75 10 N/C (0.060 m) E- × = × ⋅ = × (c) 0.110 m r = and 6 9 2 2 7 2 15.0 10 C (8.99 10 N m /C ) 1.11 10 N/C (0.110 m) E- × = × ⋅ = × E VALUATE : Outside the shell the electric field is the same as if all the charge were concentrated at the center of the shell. But inside the shell the field is not the same as for a point charge at the center of the shell, inside the shell the electric field is zero. 22.13. (a) I DENTIFY and S ET U P : It is rather difficult to calculate the flux directly from d Φ = ⋅ E A r r ú since the magnitude of E r and its angle with d A r varies over the surface of the cube. A much easier approach is to use Gauss's law to calculate the total flux through the cube. Let the cube be the Gaussian surface. The charge enclosed is the point charge. E XECUTE : 6 6 2 encl 12 2 2 9.60 10 C / 1.084 10 N m /C. 8.854 10 C / N m E Q-- × Φ = = = × ⋅ × ⋅ P By symmetry the flux is the same through each of the six faces, so the flux through one face is ( 29 6 2 5 2 1 6 1.084 10 N m /C 1.81 10 N m /C. × ⋅ = × ⋅ (b) E VALUATE : In part (a) the size of the cube did not enter into the calculations. The flux through one face depends only on the amount of charge at the center of the cube. So the answer to (a) would not change if the size of the cube were changed....
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Chapter 22 HW Solutions - PHYS 241 Ch 22 Homework Solutions...

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