Physics Solution Manual for 1100 and 2101

11b we have that 1c v2 2 22 1 c1v12 2 2 where v2 is

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Unformatted text preview: s because the electric potential V1 produced by the other two charges at corner 1 is zero, since they are infinitely far away. Once the 8.00-µC charge is in place, the electric potential V2 that it creates at corner 2 is V2 = kq1 r21 where r21 = 5.00 m is the distance between corners 1 and 2, and q1 = 8.00 µC. When the 20.0-µC charge is placed at corner 2, its electric potential energy EPE2 is kq EPE 2 = q2V2 = q2 1 r 21 ( = 20.0 × 10 −6 ( )( ) 8.99 × 109 N ⋅ m 2 /C2 8.00 × 10−6 C = 0.288 J C 5.00 m ) The electric potential V3 at the remaining empty corner is the sum of the potentials due to the two charges that are already in place on corners 1 and 2: V3 = kq1 r31 + kq2 r32 where q1 = 8.00 µC, r31 = 3.00 m, q2 = 20.0 µC, and r32 = 4.00 m. When the third charge (q3 = −15.0 µC) is placed at corner 3, its electric potential energy EPE3 is Chapter 19 Problems 1029 kq q kq q EPE3 = q3V3 = q3 1 + 2 = q3 k 1 + 2 r r 31 r32 31 r32 8.00 × 10−6 C 20.0 × 10−6 C = −15.0 × 10−6 C 8.99 × 109 N ⋅ m 2 /C2 + = − 1.034 J 3.00 m 4.00 m ( )( ) The electric potential energy of the entire array is given by EPE = EPE1 + EPE 2 + EPE3 = 0 + 0.288 J + ( − 1.034 J) = −0.746 J 28. REASONING AND SOLUTION The electrical potential energy of the group of charges is EPE = kq1q2/d + kq1q3/(2d) + kq2q3/d = 0 so q1q2 + (1/2)q1q3 + q2q3 = 0 a. If q1 = q2 = q, then q + (1/2)q3 + q3 = 0 or q3 = − 2 q 3 b. If q1 = q and q2 = −q then −q + (1/2)q3 − q3 = 0 or q3 = − 2q 29. REASONING The only force acting on each particle is the conservative electric force. Therefore, the total energy (kinetic energy plus electric potential energy) is conserved as the particles move apart. In addition, the net external force acting on the system of two particles is zero (the electric force that each particle exerts on the other is an internal force). Thus, the total linear momentum of the system is also conserved. We will use the conservation of energy and the conservation of linear momentum to find the final speed of each particle. SOLUTION For two points, A and B, along the motion, the conservation of energy is 1 mv 2 + 1 mv 2 1, A 2 1442444 4 2 2, A 3 Initial kinetic energy of the two particles + kq1q2 rA { Initial electric potential energy = 1 mv 2 + 1 mv 2 1, B 2 1442444 4 2 2, B 3 Final kinetic energy of the two particles + kq1q2 rB { Final electric potential energy Setting v1,A = v2,A = 0 since the particles are initially at rest, and letting rB = 1 rA , the 3 conservation of energy equation becomes 1030 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 1 mv 2 1, B 2 2 + 1 mv2,B = − 2 2kq1q2 rA (1) This equation cannot be solved for v1,B because the final speed v2,B of the second particle is not known. To find this speed, we will use the conservation of linear momentum: mv1, A + mv2, A = 144 244 3 Initial linear momentum mv1, B + mv2,B 14 244 4 3 Final linear momentum Setting v1,A = v2,A = 0 and solving for v2,B gives v2,B = − v1,B. Substituting this result into Equation (1) and solving for v1,B yields v1, B = = −2k q1q2 m rA ( )( )( −2 9.0 × 109 N ⋅ m 2 / C2 +5.0 × 10−6 C −5.0 × 10−6 C ( 6.0 × 10−3 kg ) ( 0.80 m ) ) = 9.7 m/s This is also the speed of v2,B. 30. REASONING AND SOLUTION The information about the electric field requires that k q2 /(1.00 m)2 = k q1 /(4.00 m)2 so q2 = (1/16.0) q1 Since q2 is negative and q1 is positive, this result implies that −q2 = q1 /16.0 (1) Let x be the distance of the zero-potential point from the negative charge and d be the separation between the charges. Then, the total potential is kq1/(d + x) + kq2/x = 0 if the point is to the right of q2 and kq1/(d − x) + kq2/x = 0 if the point is between the charges and to the left of q2. Using Equation (1) and solving for x in each case gives Chapter 19 Problems 1031 x = 0.200 m to the right of the negative charge x = 0.176 m to the left of the negative charge 31. REASONING The electric potential V at a distance r from a point charge q is V = kq/r (Equation 19.6). The potential is the same at all points on a spherical surface whose distance from the charge is r =kq/V. We will use this relation to find the distance between the two equipotential surfaces. SOLUTION The radial distance r75 from the charge to the 75.0-V equipotential surface is r75 = kq/V75, and the distance to the 190-V equipotential surface is r190 = kq/V190. The distance between these two surfaces is r75 − r190 = 1 kq kq 1 − = kq − V V75 V190 75 V190 N ⋅ m2 1 1 −8 = 8.99 × 109 2 +1.50 ×10 C 75.0 V − 190 V = 1.1 m C ( ) 32. REASONING The equipotential surfaces that surround a point charge q are spherical. kq (Equation 19.6), where Each has a potential V that is given by V = r k = 8.99 × 109 N⋅m2/C2 and r is the radius of the sphere. Equation 19.6 can be solved for q, provided that we can obtain a value for the radius r. The radius can be found from the given surface area A, since the surface area of a sphere is A = 4π r 2 . SOL...
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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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