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Physics Solution Manual for 1100 and 2101

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Unformatted text preview: 50 m 0.25 m 1018 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 15. SSM REASONING The potential of each charge q at a distance r away is given by Equation 19.6 as V = kq/r. By applying this expression to each charge, we will be able to find the desired ratio, because the distances are given for each charge. SOLUTION According to Equation 19.6, the potentials of each charge are VA = kqA rA and VB = kqB rB Since we know that VA = VB, it follows that kqA kqB = rA rB or qB rB 0.43 m = = = 2.4 qA rA 0.18 m 16. REASONING The electric potential energy EPE of the third charge q3 is given by EPE = q3VTotal (see Equation 19.3), where VTotal is the total potential at the point where the third charge is placed. The total potential at either of the empty corners is the sum of the individual potentials created by each of the charges. The individual potential created by a point charge q is V = kq/r (Equation 19.6), where r is the distance between the charge and an empty corner. SOLUTION We begin by noting that each side of the square has a length L, so that the diagonal has a length of 2 L , according to the Pythagorean theorem. Using Equation 19.6, we can express the total potential at corners A and B as follows: VTotal, A = kq2 kq1 + L 2L and VTotal, B = kq1 kq2 + L 2L Using Equation 19.3, we find that the electric potential energy of the third charge at each corner is: Chapter 19 Problems 1019 kq q k q kq EPE A = q3VTotal, A = q3 2 + 1 = 3 q2 + 1 L 2L 2 L ( −6.0 × 10−9 C )(8.99 × 109 N ⋅ m2 /C2 ) +4.0 × 10−9 C + +1.5 × 10−9 C = 0.25 m 2 = −1.1 × 10−6 J kq q k q kq EPE B = q3VTotal, B = q3 1 + 2 = 3 q1 + 2 L 2L 2 L = ( −6.0 × 10−9 C ) (8.99 × 109 N ⋅ m2 /C2 ) +1.5 × 10−9 C + +4.0 × 10−9 C 0.25 m 2 = −0.93 × 10−6 J 17. SSM WWW REASONING Initially, suppose that one charge is at C and the other charge is held fixed at B. The charge at C is then moved to position A. According to Equation 19.4, the work WCA done by the electric force as the charge moves from C to A is WCA = q (VC − VA ) , where, from Equation 19.6, VC = kq / d and VA = kq / r . From the figure at the right we see that d = r 2 + r 2 = 2r . Therefore, we find that d r B kq kq kq 2 1 WCA = q – = – 1 r 2 2r r SOLUTION Substituting values, we obtain WCA = C A (8.99 ×109 N ⋅ m 2 / C2 )(3.0 ×10−6 C) 2 1 – 1 = –4.7 ×10−2 J 0.500 m 2 r 1020 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 18. REASONING The electric potential at a distance r from a point charge q is given by Equation 19.6 as V = kq / r . The total electric potential at location P due to the four point charges is the algebraic sum of the individual potentials. q d SOLUTION The total electric potential at P is (see the drawing) V= k ( −q ) d + k ( +q ) 2d + k ( +q ) d + k ( −q ) d = −q q d d P − kq 2d d −q Substituting in the numbers gives N ⋅ m2 − 8.99 ×109 2.0 × 10−6 C 2 C − kq = = −9.4 ×103 V V= 2d 2 ( 0.96 m ) ( ) 19. REASONING The electric potential at a distance r from a point charge q is given by Equation 19.6 as V = kq / r . The total electric potential at location P due to the six point charges is the algebraic sum of the individual potentials. +7.0q +5.0q +3.0q d d d d P d d −5.0q −3.0q +7.0q SOLUTION Starting at the upper left corner of the rectangle, we proceed clockwise and add up the six contributions to the total electric potential at P (see the drawing): V= k ( +7.0q ) d d + 2 2 2 = k ( +14.0q ) d d + 2 2 2 + k ( +3.0q ) k ( +5.0q ) k ( +7.0q ) k ( −3.0q ) k ( −5.0q ) + + + + d d 2 2 2 2 d 2 d 2 d d + d + d + 2 2 2 2 2 Chapter 19 Problems Substituting q = 9.0 × 10 −6 1021 C and d = 0.13 m gives 2 9 N⋅m +14.0 ) 9.0 × 10−6 C 8.99 ×10 2 ( k ( +14.0q ) C = = +7.8 × 106 V V= 2 2 d 0.13 m d2 + ( 0.13 m )2 + 2 2 ( ) 20. REASONING The potential from the positive charge is positive, while the potential from the negative charge is negative at the spot in question. The magnitude of the positive potential must be equal to the magnitude of the negative potential. Only then will the algebraic sum of the two potentials give zero for the total potential. The drawing shows the arrangement of the charges. the two charges and the spot in question form a right triangle. Therefore, the distance between the positive charge and the zero-potential spot is the hypotenuse of the right triangle and, according to the Pythagorean theorem, is greater than L. Zero potential L –q +2q 2.00 m There are two spots on the dashed line where the total potential is zero. One spot is shown in the drawing. The other spot is on the dashed line at a distance L below the negative charge. SOLUTION We begin by noting that the distance between the positive charge +2q and the zero-potential spot is given by the Pythagorean theorem as L2 + ( 2.00 m ) . With this in 2 mind and using V = kq/r (Equation 19.6), we write the total potential at the spot in question as follows: VTotal = k ( −q ) k ( +2 q ) + =0 2 L L2 + ( 2.00 m ) or +2 1 = L L2 + ( 2.00 m ) Squaring both...
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