# Chapt23 - Chapter 23 Gauss Law Problem 1 The square surface shown in Fig 23-26 measures 3.2 mm on each side It is immersed in a uniform electric

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Chapter 23: Gauss' Law Problem 1 The square surface shown in Fig. 23-26 measures 3.2 mm on each side. It is immersed in a uniform electric field with E = 1800 N/C. The field lines make an angle of 35 ° with a normal to the surface, as shown. Take the normal to be outward as if the surface were one face of a box. Calculate the electric flux through the surface. Normal 35 o Figure 23-26 Answer - 1.51 × 10 -2 N-m 2 /C. Solution Flux is defined to be Φ = E d A = E cos ( θ ) d A . In this case E and θ are constant over the area so that Φ = E cos( θ ) d A = EA cos( θ ), where A is the area of the square. Thus Φ = (1800 N/C)*(3.2 mm) 2 cos(145 ° ), (Note that the angle is 145 ° ) Φ = - 1.51 × 10 -2 N-m 2 /C. Problem 3 The cube in Fig. 23-27 has edge lengths of 1.40 m and is oriented as shown in a region of uniform electric field. Find the electric flux through the right face if the electric field, in newtons per coulomb, is given by (a) 6.00 i , (b) –2.00 j , and (c) –3.00 i + 4.00 k . (d) What is the total flux through the cube for each of these fields? Answer (a) zero. (b) -3.92 N-m 2 /C. (c) zero. (d) zero for each field. Figure 23-27

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23-2 Chapter 23: Gauss’ Law Solution The vector area of the right face is A = (1.4 m) 2 j = 1.96 m 2 j . Since E is a constant in each of the cases the flux is just Φ = E A . (a) Here E = 6 N/C i . Therefore Φ = E A Φ = (6 N/C i ) (1.96 m 2 j ) = 0. (b) Here E = -2 N/C j . Therefore Φ = E A Φ = (-2 N/C j ) - (1.96 m 2 j ) = -3.92 N-m 2 /C. (c) Here E = -3 N/C i + 4 N/C k . Therefore Φ = E A Φ = (-3 N/C i + 4 N/C k ) (1.96 m 2 j ) = 0. (d) The answer is zero in each case. The total flux is just the sum of the flux for each of the six faces Φ tot = Φ top + Φ bottom + Φ left + Φ right + Φ front + Φ back Φ tot = E A + E A bottom + E A left + E A right + E A front + E A Φ tot = E ( A + A bottom + A left + A right + A front + A ) This last step is justified since E is the same on every face of the cube. The sum of the vector areas is zero (all have same size and they occur in pairs of anti-parallel vectors). Problem 5 A point charge of 1.8 μ C is at the center of a cubical Gaussian surface 55 cm on edge. What is Φ through the surface? Answer 2.03 × 10 5 N-m 2 /C. From Gauss' law the flux through a closed surface is Φ = E d A = Q/ ε o , where Q is the net charge inside the Gaussian surface. If you know the charge inside the surface it is not necessary to know about the geometry of the surface in order to calculate the flux - it is just Q/ ε o . Therefore the flux through the cube is just
Chapter 23: Gauss’ Law 23-3 Φ = Q/ ε o Φ = (1.8 μ C)/ ε o = 2.03 × 10 5 N-m 2 /C. Notice that we are not assuming that the electric field is constant on the surface or anything else about the field. This solution only tells us about the flux. Problem 9 It is found experimentally that the electric field in a certain region of the Earth's atmosphere is directed vertically down. At an altitude of 300 m the field has magnitude 60.0 N/C; at an altitude of 200 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 100 m on edge, with horizontal faces at altitudes of 200 and 300 m.

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## This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.

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Chapt23 - Chapter 23 Gauss Law Problem 1 The square surface shown in Fig 23-26 measures 3.2 mm on each side It is immersed in a uniform electric

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