HW3%20solutions

HW3%20solutions - Physics 204 morning summer 2008 Homework...

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Unformatted text preview: Physics 204 morning summer 2008 Homework 3 solutions Chapter 19 Problems 7,3,19,18,54 7. SSM REASONING The only force acting on the moving electron is the conservative electric force. Therefore, the total energy of the electron (the sum of the kinetic energy KE and the electric potential energy EPE) remains constant throughout the trajectory of the electron. Let the subscripts A and B refer to the initial and final positions, respectively, of the electron. Then, 1 mv 2 A 2 Solving for vB gives 2 vB = vA – 2 + EPE A = 1 mvB + EPE B 2 2 (EPE B –EPE A ) m Since the electron starts from rest, vA = 0 m/s. The difference in potential energies is related to the difference in potentials by Equation 19.4, EPE B − EPE A = q (VB − VA ) . SOLUTION The speed vB of the electron just before it reaches the screen is vB = – 2q 2(–1.6 ×10−19 C) (VB –VA ) = – (25 000 V) = 9.4 ×107 m/s −31 m 9.11× 10 kg 3. REASONING AND SOLUTION The only force that acts on the ‐particle is the conservative electric force. Therefore, the total energy of the ‐particle is conserved as it moves from point A to point B: 1 mv 2 + EPE = 1 mv 2 + EPE A A B B 2 2 Total energy at point A Total energy at point B Since the ‐particle starts from rest, vA = 0 m/s. The electric potential V is related to the electric potential energy EPE by V = EPE/q (see Equation 19.3). With these changes, the equation above gives the kinetic energy of the ‐ particle at point B to be 1 mv 2 B 2 = EPE A − EPE B = q (VA − VB ) Since an ‐particle contains two protons, its charge is q = 2e = 3.2 10 volts) is 1 mv 2 B 2 ( –19 C. Thus, the kinetic energy (in electron‐ ) = q (VA − VB ) = 3.2 × 10−19 C ⎡ +250 V − ( −150 V ) ⎤ ⎣ ⎦ = 1.28 × 10−16 ⎛ 1.0 eV J⎜ −19 ⎝ 1.6 × 10 ⎞ 2 ⎟ = 8.0 × 10 eV J⎠ 19. SSM REASONING The only force acting on the moving charge is the conservative electric force. Therefore, the sum of the kinetic energy KE and the electric potential energy EPE is the same at points A and B: 1 mv 2 A 2 2 + EPE A = 1 mvB + EPE B 2 Since the particle comes to rest at B, v B = 0 . Combining Equations 19.3 and 19.6, we have ⎛ kq ⎞ EPE A = qVA = q ⎜ 1 ⎟ ⎝ d ⎠ ⎛ kq ⎞ EPE B = qVB = q ⎜ 1 ⎟ ⎝ r ⎠ and where d is the initial distance between the fixed charge and the moving charged particle, and r is the distance between the charged particles after the moving charge has stopped. Therefore, the expression for the conservation of energy becomes 1 mv 2 A 2 + kqq1 kqq1 = d r This expression can be solved for r. Once r is known, the distance that the charged particle moves can be determined. SOLUTION Solving the expression above for r gives r= = kqq1 kqq1 1 mv 2 + A 2 d (8.99 ×109 N ⋅ m 2 / C2 )(–8.00 × 10− 6 C)(−3.00 ×10− 6 C) 9 2 2 −6 −6 –3 2 (8.99 × 10 N ⋅ m / C )( – 8.00 × 10 C)( −3.00 × 10 C) 1 (7.20 × 10 kg)(65.0 m/s) + 2 0.0450 m = 0.0108 m Therefore, the charge moves a distance of 0.0450 m – 0.0108 m = 0.0342 m . 18. REASONING The potential V created by a point charge q at a spot that is located at a distance r is given by Equation 19.6 as V = kq , where q can be either a positive or negative quantity, depending on the nature of the charge. We r will apply this expression to obtain the potential created at the empty corner by each of the three charges fixed to the square. The total potential at the empty corner is the sum of these three contributions. Setting this sum equal to zero will allow us to obtain the unknown charge. SOLUTION The drawing at the right shows the three charges fixed to the corners of the square. The length of each side of the +1.8 μC square is denoted by L. Note that the distance between the unknown charge q and the empty corner is L. Note also that the L +1.8 μC L L distance between one of the 1.8‐μC charges and the empty corner is r = L , but that the distance between the other 1.8‐μC charge and the empty corner is r = L2 + L2 = 2 L , according to the Pythagorean theorem. Using Equation 19.6 to express the potential created by the unknown charge q and by each of the 1.8‐μC charges, we find that the total potential at the empty corner is Vtotal ( ) ( ) −6 −6 kq k +1.8 × 10 C k +1.8 × 10 C = + + = 0 L L 2L In this result the constant k and the length L can be eliminated algebraically, leading to the following result for q: q + 1.8 × 10 −6 1.8 ×10−6 C C+ =0 2 or ( ) 1 ⎞ ⎛ −6 q = −1.8 ×10−6 C ⎜ 1 + ⎟ = −3.1× 10 C 2⎠ ⎝ 54. REASONING The electric potential at a distance r from a point given by Equation 19.6 as V = kq / r . The total electric location P due to the four point charges is the algebraic sum individual potentials. SOLUTION The total electric potential at P is (see the V= k ( −q ) d + k ( +q ) 2d + k ( +q ) d + k ( −q ) d = charge q is potential at of the q d −q q drawing) d d −kq 2d P d Substituting in the numbers gives −q ⎛ N ⋅ m2 ⎞ 2.0 ×10−6 C − ⎜ 8.99 × 109 2 ⎟ C ⎠ −kq V= = ⎝ = −9.4 × 103 V 2d 2 ( 0.96 m ) ( ) ...
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This note was uploaded on 10/04/2009 for the course PHYS 750:204 taught by Professor Rollino during the Spring '09 term at Rutgers.

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