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Unformatted text preview: CHAPTER 24 GAUSSS LAW ActivPhysics can help with these problems: Activities 11.411.6 Section 24 1: Electric Field Lines Problem 1. What is the net charge shown in Fig. 2439? The magnitude of the middle charge is 3 C m . Solution The number of lines of force emanating from (or terminating on) the positive (or negative) charges is the same (14 in Fig. 2439), so the middle charge is  3 C m and the outer ones are + 3 C m . The net charge shown is therefore 3 3 3 3 + = C. m This is reflected by the fact that 14 lines emerge from the boundary of the figure. FIGURE 2439 Problem 1 Solution. Problem 2. A charge + 2 q and a charge  q are near each other. Sketch some field lines for this charge distribution, using the convention of eight lines for a charge of magnitude q . Solution The sketch is similar to Fig. 244( ) f with twice the number of lines of force. Problem 3. Two charges + q and a charge  q are at the vertices of an equilateral triangle. Sketch some field lines for this charge distribution. Solution (The sketch shown follows the texts convention of eight lines of force per charge magnitude q .) CHAPTER 24 569 Problem 3 Solution. Problem 4. The net charge shown in Fig. 2440 is + Q . Identify each of the charges A , B , C shown. FIGURE 2440 Problem 4. Solution 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. Thus, Q Q Q C B A =  = 2 . The total charge is Q Q Q Q Q A B C A = + + = , so Q Q Q C B = =  2 . Section 242: Electric Flux Problem 5. A flat surface with area 2 0 2 . m is in a uniform electric field of 850 N C / . What is the electric flux through the surface when it is (a) at right angles to the field, (b) at 45 to the field, and (c) parallel to the field? Solution (a) When the surface is perpendicular to the field, its normal is either parallel or antiparallel to E . Then Equation 241 gives = = = = E A EA cos( ) ( / )( ) . / . 180 850 2 1 70 2 2 or N C m kN m C (b) = = = E A EA cos( ) 45 135 or = ( . / )( . ) . / . 1 70 0 866 1 20 2 2 kN m C kN m C (c) = = EA cos . 90 570 CHAPTER 24 Problem 6. What is the electric field strength in a region where the flux through 1 0 1 0 . . cm cm flat surface is 65 2 N m C / , if the field is uniform and the surface is at right angles to the field? Solution The magnitude of the flux through a flat surface perpendicular to a uniform field is = EA (see solution to part (a) of the previous problem). Thus E A = = = = = ( / ( ) / . 65 10 650 2 4 2 N m C m kN C Problem 7. A flat surface with area 0 14 2 . m lies in the x y plane, in a uniform electric field given by E = + + 51 2 1 3 5 . ....
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This note was uploaded on 10/19/2010 for the course PHY 78647 taught by Professor Miller during the Spring '09 term at Albany College of Pharmacy and Health Sciences.
 Spring '09
 Miller
 Physics, Charge, Force

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