Physics Solution Manual for 1100 and 2101

99 109 n m2 c2 12 106 c 45 106 c 30 m 2 f23 3 45 x

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Unformatted text preview: ual magnitudes and cancel, leaving a resultant that is parallel to the x axis. 10. 8.5 µ C 11. (e) According to Equation 18.2, the force exerted on a charge by an electric field is proportional to the magnitude of the charge. Since the negative charge has twice the magnitude of the positive charge, the negative charge experiences twice the force. Furthermore, the direction of the force on the positive charge is in the same direction as the field, so that we can conclude that the field points due west. The force on the negative charge points opposite to the field and, therefore, points due east. 944 ELECTRIC FORCES AND ELECTRIC FIELDS 12. (c) The electric field created by a point charge has a magnitude E = kq and is inversely r2 proportional to the square of the distance r. If r doubles, the charge magnitude must increase by a factor of 22 = 4 to keep the field the same. 13. (b) To the left of the positive charge the two contributions to the total field have opposite directions. There is a spot in this region at which the field from the smaller, but closer, positive charge exactly offsets the field from the greater, but more distant, negative charge. 14. (e) Consider the charges on opposite corners. In all of the arrangements these are like charges. This means that the two field contributions created at the center of the square point in opposite directions and, therefore, cancel. Thus, only the charge opposite the empty corner determines the magnitude of the net field at the center of the square. Since the point charges all have the same magnitude, the net field there has the same magnitude in each arrangement. 15. 1.8 × 10−6 C/m2 16. (c) The tangent to the field line gives the direction of the electric field at a point. At A the tangent points due south, at B southeast, and at C due east. 17. (a) The electric field has a greater magnitude where the field lines are closer together. They are closest together at B and farthest apart at A. Therefore, the field has the greatest magnitude at B and the smallest magnitude at A. 18. (d) In a conductor electric charges can readily move in response to an electric field. In A, B, and C the electric charges experience an electric field and, hence, a force from neighboring charges and will move outward, away from each other. They will rearrange themselves so that the electric field within the metal is zero at equilibrium. This means that they will reside on the outermost surface. Thus, only D could represent charges in equilibrium. 19. 1.3 N·m2/C 20. 0.45 N·m2/C Chapter 18 Problems 945 CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS PROBLEMS ______________________________________________________________________________ 1. SSM REASONING AND SOLUTION The total charge of the electrons is q = N(–e) = (6.0 × 1013)(–1.60 × 10–19 C) q = – 9.6 × 10–6 C = –9.6 µC The net charge on the sphere is, therefore, qnet = +8.0 µC – 9.6 µC = −1.6 µC ______________________________________________________________________________ 2. REASONING The charge of a single proton is +e, and the charge of a single electron is –e, where e = 1.60×10−19 C. The net charge of the ionized atom is the sum of the charges of its constituent protons and electrons. SOLUTION The ionized atom has 26 protons and 7 electrons, so its net electric charge q is q = 26 ( +e ) + 7 ( −e ) = +19e = +19 (1.60 ×10−19 C ) = +3.04 ×10−18 C 3. REASONING a. Since the objects are metallic and identical, the charges on each combine and produce a net charge that is shared equally by each object. Thus, each object ends up with one-fourth of the net charge. b. The number of electrons (or protons) that make up the final charge on each object is equal to the final charge divided by the charge of an electron (or proton). SOLUTION a. The net charge is the algebraic sum of the individual charges. The charge q on each object after contact and separation is one-fourth the net charge, or q= 1 4 (1.6 µ C + 6.2 µ C – 4.8 µ C – 9.4 µ C ) = –1.6 µ C ELECTRIC FORCES AND ELECTRIC FIELDS 946 b. Since the charge on each object is negative, the charge is comprised of electrons. The number of electrons on each object is the charge q divided by the charge −e of a single electron: q −1.6 ×10−6 C = = 1.0 ×1013 Number of electrons = −19 −e −1.60 × 10 C ____________________________________________________________________________________________ 4. REASONING When N electrons, each carrying a charge −e = −1.6×10−19 C, are transferred from the plate to the rod, the system consisting of the plate and the rod is isolated. Therefore, the total charge q1i+ q2i of the system is unchanged by the process, where q1i is the initial charge of the plate and q2i is the initial charge of the rod. At the end, the rod and the plate each have the same final charge q1f = q2f. Therefore, each must have a charge equal to half the total charge of the system: q1f = q2f = 1 ( q1i + q2i ) . 2 SOLUTION The final charge q2f on the rod is equ...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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