Homework 3

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Southern New Hampshire University *

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Apr 3, 2024

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Homework #3 25. A square surface of area 2cm2 is in a space of uniform electric field of magnitude 103N/C. The amount of flux through it depends on how the square is oriented relative to the direction of the electric field. Find the electric flux through the square, when the normal to it makes the following angles with electric field: (a) 30°, (b) 90°, and (c) 0°. Note that these angles can also be given as 180°+ θ . A. (2x10^-2Nm^2/C)(Square root(3))/2 =(1.732)(10^-1Nm^2/C) =0.1732 Nm^2/C B. (10^3 N/C)(2x10^-4m^2cos(90)) (2x10^-1Nm^2/C)(0) =0 C. (10^3N/C)(2x10^4m^2)(cos0) =(2x10^-1Nm^2/C)(1) =0.2Nm^2/C 31. Find the electric flux through the closed surface whose cross-sections are shown below. 38. A cube whose sides are of length d is placed in a uniform electric field of magnitude E=4.0×103N/C so that the field is perpendicular to two opposite faces of the cube. What is the net flux through the cube? The flux through the cube would be 0. 43. A very long, thin wire has a uniform linear charge density of 50 μ C/m. What is the electric field at a distance 2.0 cm from the wire? E=(50UC/m)(10^-6/U)/2(3.14)(8.85x10^-12F/m)(2cm)(m/100cm) =4.5x10^7N/C

44. A charge of −30μ C is distributed uniformly throughout a spherical volume of radius 10.0 cm. Determine the electric field due to this charge at a distance of (a) 2.0 cm, (b) 5.0 cm, and (c) 20.0 cm from the center of the sphere. a. E=9x10^9x-30x10^-6x2x10^-2/10^3x10^-6 =-5.4 x 10^6N/c b. E= 9x10^9x-30x10^-2x5^-2/10^3x10^-6 =-13.5 x 10^6 N/C c. 9x10^9x-30x10^-6x20x10^-2/0.10^3 =-6.75 x 10^-6N/C 52. A long copper cylindrical shell of inner radius 2 cm and outer radius 3 cm surrounds concentrically a charged long aluminum rod of radius 1 cm with a charge density of 4 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod: (a) 0.5 cm, (b) 1.5 cm, (c) 2.5 cm, (d) 3.5 cm, and (e) 7 cm. a. 0.5cm from the aluminum rod is 0 b. (4)(10^-12/1)/2(3.14)(1.5)(10^-2)/(cm)) =4.8rN/m c. 2.5cm from the aluminum rod is 0 d. (4)(10^-12/1)/2(3.14)(8.85x10^-12)((3.5)(10^-2)/cm)) =2.05r N/m e. (4)/2(3.14)(8.85x10^-12)(7)(10^-2)(cm) =1.025r N/m

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