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21_InstSolManual_PC - GAUSS'S LAW 21 EXERCISES Section 21.1...

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GAUSS’S LAW E XERCISES Section 21.1 Electric Field Lines 18. The number of lines of force emanating from (or terminating on) the positive (or negative) charges is the same (14 in Fig. 21.31), so the middle charge is 3 C μ - and the outer ones are 3 C. μ + The net charge shown is therefore 3 3 3 3 C. μ + - = This is reflected by the fact that 14 lines emerge from the boundary of the figure. 19. I NTERPRET This problem is about drawing field lines to represent the field strength of a charge configuration. D EVELOP We follow the methodology illustrated in Figure 21.3. There are 16 lines emanating from charge +2 q (eight for each unit of + q ). Similarly, we have 8 lines ending on . q - E VALUATE The field lines of the charge configuration are shown below. A SSESS Our sketch is similar to Fig. 21.3 ( f ) with twice the number of lines of force. 20. (The sketch shown follows the text’s convention of eight lines of force per charge magnitude q .) 21. I NTERPRET In this problem we are asked to identify the charges based on the pattern of the field lines. D EVELOP From the direction of the lines of force (away from positive and toward negative charge) one sees that A and C are positive and B is a negative charge. Eight lines of force terminate on B , eight originate on C , but only four originate on A , so the magnitudes of B and C are equal, while the magnitude of A is half that value. E VALUATE Based on the reasoning above, we may write 2 . C B A Q Q Q = - = The total charge is A Q Q = + , B C A Q Q Q + = so . 2 C B Q Q Q = = - A SSESS The magnitude of the charge is proportional to the number of field lines emerging from or terminating at the charge. 21.1 21
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21.2 Chapter 21 Section 21.2 Electric Flux and Field 22. (a) When the surface is perpendicular to the field, its normal is either parallel or anti-parallel to . E r Then Equation 21.1 gives 2 2 cos(0 or 180 ) (850 N/C)(2 m ) 1.70 kN m /C. E A EA Φ = = ° ° = ± = ± r r (b) cos(45 or 135 ) E A EA Φ = = ° ° = r r 2 2 (1.70 kN m /C)(0.866) 1.20 kN m /C. ± = ± (c) cos90 0. EA Φ = ° = 23. I NTERPRET This problem is about finding the electric field strength, given the flux through a surface. D EVELOP The magnitude of the flux through a flat surface perpendicular to a uniform field is | | EA Φ = E VALUATE From the equation above, we find the field strength to be 2 4 2 | | 65 N m /C 650 kN/C 10 m E A - Φ = = = A SSESS The general expression for the electric flux Φ is given by Equation 21.1: cos , E A EA θ Φ = = r r where θ is the angle between the normal vector A r and the electric field . E r We see that when 0, θ = EA Φ = is a maximum, and when 180 , EA θ = ° Φ = - is a minimum. 24. The surface can be represented by a vector area 2 ˆ (0.14 m )( ). A k = ± r (Since the surface is open, we have a choice of normal to the x - y plane.) Then 2 2 2 ˆ (0.14 m ) (0.14 m ) (3.5 kN/C)(0.14 m ) z E A E k E Φ = = ± × = ± = ± = r r r 2 490 N m /C. ± (Only the z component of the field contributes to the flux through the x - y plane.) 25. I NTERPRET This problem is about the electric flux through the surface of a sphere.
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