Physics Solution Manual for 1100 and 2101

00 102 kg 300 ms 200 ms 400 105 c 7850

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Unformatted text preview: ave that r Chapter 19 Problems kq 2kq VA = 2 = r r A A 1025 kq 2kq VB = 2 = r r B B and (2) In Equation (2), rA = 0.480 m is the length of one side of the square, and rB is half the length ( ) of the diagonal rA 2 of the square, so that rB = 1 rA 2 = 1 2 ( 0.480 m ) = 0.339 m . 2 2 Substituting Equations (2) into EPE = q0V (Equation 19.3) yields 2kq 2kq0 q EPE A = q0VA = q0 r = r A A 2kq 2kq0 q EPE B = q0VB = q0 r = r B B and (3) Substituting Equations (3) into Equation (1), we obtain 1 2 2 mvB + 2 kq0 q rB = 2 kq0 q (4) rA Solving Equation (4) for vB2, we find that 1 mv 2 B 2 = 2 kq0 q rA − 2kq0 q rB or 1 mv 2 B 2 1 1 = 2kq0 q − r A rB or 2 vB = 4kqq0 1 1 − m rA rB Therefore, the speed of the test charge at the center of the square is vB = = 4kqq0 1 1 − m rA rB ( )( 7.20 ×10−6 C )( −2.40 ×10−8 C ) 1 − 1 0.480 m 0.339 m ( 6.60 ×10−8 kg ) 4 8.99 × 109 N ⋅ m 2 /C 2 = 290 m/s 1026 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 25. REASONING The only force acting on each proton is the conservative electric force. Therefore, the total energy (kinetic energies plus electric potential energy) is conserved at all points along the motion. For two points, A and B, the conservation of energy is expressed as follows: 2 2 1 mv 2 + 1 mv 2 + EPE = 1 mvB + 1 mvB + EPE B A A 2 2 2 2 1 24 4A 3 1442443 1 24 43 1442443 Initial kinetic energies of the two protons Initial electric potential energy Final kinetic energies of the two protons Final electric potential energy The electric potential energy of two protons (charge on each proton = +e) that are separated by a distance r can be found by combining the relations EPE = eV (Equation 19.3) and 2 V = ke/r (Equation 19.6) to give EPE = ke /r. By using this expression for EPE in the conservation of energy relation, we will be able to determine the distance of closest approach. SOLUTION When the protons are very far apart (rA = ∞), so that EPEA = 0 J. At the distance rB of closest approach, the speed of each proton is momentarily zero (vB = 0 m/s). With these substitutions, the conservation of energy equation reduces to 1 mv 2 A 2 + 1 mv 2 = A 2 k e2 rB Solving for rB, the distance of closest approach, gives ( )( 8.99 × 109 N ⋅ m 2 / C 2 1.60 × 10−19 C ke2 rB = = 2 2 mv A 1.67 × 10−27 kg 3.00 × 106 m/s ( )( ) ) 2 = 1.53 × 10−14 m We have taken the mass of a proton from the inside of the front cover of the text. 26. REASONING It will not matter in what order the group is assembled. For convenience, we will assemble the group from one end of the line to the other. The potential energy of each charge added to the group will be determined and the four values added together to get the total. At each step, the electric potential energy of an added charge q0 is given by Equation 19.3 as EPE = q0VTotal, where VTotal is the potential at the point where the added charge is placed. The potential VTotal will be determined by adding together the contributions from the charges previously put in position, each according to Equation 19.6 (V = kq/r). Chapter 19 Problems 1027 SOLUTION The first charge added to the group has no electric potential energy, since the spot where it goes has a total potential of VTotal = 0 V, there being no charges in the vicinity to create it: ( ) EPE1 = q0VTotal = 2.0 ×10−6 C ( 0 V ) = 0 J The second charge experiences a total potential that is created by the first charge: VTotal ( )( ) 8.99 × 109 N ⋅ m 2 / C2 +2.0 × 10−6 C kq = = = 4.5 × 104 V r 0.40 m ( )( ) EPE 2 = q0VTotal = 2.0 × 10−6 C 4.5 × 104 V = 0.090 J The third charge experiences a total potential that is created by the first and the second charges: VTotal (8.99 × 109 N ⋅ m2 / C2 ) ( +2.0 × 10−6 C) = 0.80 m (8.99 × 109 N ⋅ m2 / C2 )( +2.0 × 10−6 C ) = 6.7 × 104 V + 0.40 m ( )( ) EPE 3 = q0VTotal = 2.0 × 10−6 C 6.7 × 104 V = 0.13 J The fourth charge experiences a total potential that is created by the first, second, and third charges: VTotal (8.99 × 109 N ⋅ m2 / C2 ) ( +2.0 × 10−6 C) = 1.2 m (8.99 × 109 N ⋅ m2 / C2 )( +2.0 × 10−6 C) + 0.80 m (8.99 × 109 N ⋅ m2 / C2 )( +2.0 × 10−6 C ) = 8.2 × 104 V + 0.40 m ( )( ) EPE 4 = q0VTotal = 2.0 × 10−6 C 8.2 × 104 V = 0.16 J The total electric potential energy of the group is 1028 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL EPE Total =EPE1 + EPE 2 + EPE 3 + EPE 4 = 0 J + 0.090 J + 0.13 J + 0.16 J= 0.38 J 27. SSM REASONING Initially, the three charges are infinitely far apart. We will proceed as in Example 8 by adding charges to the triangle, one at a time, and determining the electric potential energy at each step. According to Equation 19.3, the electric potential energy EPE is the product of the charge q and the electric potential V at the spot where the charge is placed, EPE = qV. The total electric potential energy of the group is the sum of the energies of each step in assembling the group. SOLUTION Let the corners of the triangle be numbered clockwise as 1, 2 and 3, starting with the top corner. When the first charge (q1 = 8.00 µC) is placed at a corner 1, the charge has no electric potential energy, EPE1 = 0. This i...
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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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