CH24 - CHAPTER 24 GAUSS’S LAW ActivPhysics can help with...

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CHAPTER 24 GAUSS’S LAW ActivPhysics can help with these problems: Activities 11.4–11.6 Section 24 - 1: Electric Field Lines Problem 1. What is the net charge shown in Fig. 24-39? The magnitude of the middle charge is 3C µ . Solution The number of lines of force emanating from (or terminating on) the positive (or negative) charges is the same (14 in Fig. 24-39), so the middle charge is 3 C and the outer ones are + . The net charge shown is therefore 3333 +−= C. This is reflected by the fact that 14 lines emerge from the boundary of the figure. FIGURE 24-39 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. 24-4( ) 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 text’s convention of eight lines of force per charge magnitude q .)
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CHAPTER 24 569 Problem 3 Solution. Problem 4. The net charge shown in Fig. 24-40 is + Q . Identify each of the charges A , B , C shown. A B C FIGURE 24-40 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, QQQ C B A =− = 2. The total charge is QQ Q Q Q A B C A = + + = , so QQ Q C B == 2 . Section 24-2: Electric Flux Problem 5. A flat surface with area 20 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 anti-parallel to E . Then Equation 24-1 gives Φ= ⋅ = ° ° =± EA EA cos( ) ( / )( ) . / . 0 180 850 2 1 70 22 or N C m kN m C (b) Φ = = °° = EA cos( ) 45 135 or ±⋅ = ± (. / ) ( . ) . / . 170 0866 120 k NmC k NmC (c) Φ = ° = EA cos . 90 0 Problem 6. What is the electric field strength in a region where the flux through 10 10 .. cm cm × flat surface is 65 2 /, 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 Φ = EA (see solution to part (a) of the previous problem). Thus ==⋅ = Φ (/ ( ) / . 65 10 650 24 2 m k NC Problem 7. A flat surface with area 014 2 . m lies in the x - y plane, in a uniform electric field given by =++ 51 21 35 . /. j k kN C Find the flux through this surface.
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570 CHAPTER 24 Solution The surface can be represented by a vector area Ak (.
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CH24 - CHAPTER 24 GAUSS’S LAW ActivPhysics can help with...

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