Problem 22.7 From example 21.11, the electric field due to an infinite line charge is: Radius of cylinder r=0.250 m Length of cylinder l=0.400 m Infinite positive line charge λ=6.00 µC/m (a) IDENTIFY: Use Eq.(22.5) to calculate the flux through the surface of the cylinder. SET UP: The line of charge and the cylinder are sketched in Figure 22.7. g Figure 22.7 EXECUTE: The area of the curved part of the cylinder is A 2 rl. The electric field is parallel to the end caps of the cylinder, so E A 0 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 at all points on this surface. Thus 0 and E E EA cos EA / 2 P0r 2 rl l P0 6.00 10 6 C/m 0.400 m 8.854 1012 C2 / N m2 2.71105 N m 2 / C (b) In the calculation in part (a) the radius r of the cylinder divided out, so the flux remains the same, E 2.71105 N m2 / C. (c) E l P0 6.00 10 6 C/m 0.800 m 8.854 1012 C2 / N m2 5.42 105 N m2 / C (twice the flux calculated in parts (b) and (c)). EVALUATE: The flux depends on the number of field lines that pass through the surface of the cylinder. 22.13. (a) IDENTIFY and SET UP: It is rather difficult to calculate the flux directly from úE dA since the magnitude of E and its angle with dA 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. EXECUTE: E Qencl / P0 9.60 106 C 1.084 106 N m2 / C. 8.854 1012 C2 / N m2 By symmetry the flux is the same through each of the six faces, so the flux through one face is 1 6 1.084 10 6 N m2 / C 1.81105 N m2 / C. (b) EVALUATE: 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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