{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

view-notes-pcs110-002 - CHAPTER 24 GAUSSS LAW ActivPhysics...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 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. 24-39), 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 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 .)
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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. 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, 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 24-2: 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 anti-parallel to E . Then Equation 24-1 gives Φ = = ° ° = ± = ± E A EA cos( ) ( / )( ) . / . 0 180 850 2 170 2 2 or N C m kN m C (b) Φ = = ° ° = E A EA cos( ) 45 135 or ± = ± ( . / )( . ) . / . 170 0 866 120 2 2 kN m C kN m C (c) Φ = ° = EA cos . 90 0
Image of page 2
570 CHAPTER 24 Problem 6. What is the electric field strength in a region where the flux through 10 10 . . 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 Φ = EA (see solution to part (a) of the previous problem). Thus E A = = = - Φ = = ( / ( ) / . 65 10 650 2 4 2 N m C m kN C Problem 7. A flat surface with area 014 2 . m lies in the x - y plane, in a uniform electric field given by E î = + + 51 21 3 5 . . $ . $ / . j k kN C Find the flux through this surface. Solution The surface can be represented by a vector area A k = ± ( . )( $ ). 014 2 m (Since the surface is open, we have a choice of normal to the x - y plane.) Then Φ = = ± × = ± = ± = ± E A E k $ ( . ) ( . ) ( . / )( . ) / . 014 014 35 014 490 2 2 2 2 m m kN C m N m C E z (Only the z component of the field contributes to the flux through the x - y plane.) Problem 8.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern