Physics Solutions to Chapter HW Chapter : Questions:, 6, Problems:, 9, 5, 33, 35, 53, 5, 67, 99 Question - (a),, and then a tie of, 3, and 5 (where E = ); (b) negative x direction; (c) positive x direction Question -6 b, then a, c, and d tie Question - (a) Q/πε ; (b) Q/πε ; (c) Q/πε ;(d) a, b, c Problem - (a) The potential as a function of r is r r r r V ( r) = V ( ) E( r) dr= dr= πε 8πε 3 3 9 5 (899 N m C )(35 C)(5 m) = = 3 (3 m) (b) Since ΔV = V() V() = /8πε, we have ( ) V 9 5 (899 N m C )(35 C) = = = 8πε (3 m) 68 V 68 V Problem -9 irst, we observe that V (x) cannot be eual to zero for x > d In fact V (x) is always negative for x > d Now we consider the two remaining regions on the x axis: x < and < x < d (a) or < x < d we have d = x and d = d x Let V x k ( ) = + d d I I K J = = + KJ 3 x d x and solve: x = d/ With d = cm, we have x = 6 cm
(b) Similarly, for x < the separation between and a point on the x axis whose coordinate is x is given by d = x; while the corresponding separation for is d = d x We set V x k I ( ) = + d d x + 3 d x KJ = to obtain x = d/ With d = cm, we have x = cm I K J = Problem -5 (a) ll the charge is the same distance from C, so the electric potential at C is Q 6Q 5Q V = = = =3 V, π π 8 m 9 5(899 N m C )( C) ε ε where the zero was taken to be at infinity (b) ll the charge is the same distance from P That distance is potential at P is Q 6Q 5Q V = = πε + D + D πε + D = 9 5(899 N m C )( C) =78 V (8 m) + (67 m) D, + so the electric Problem -33 Consider an infinitesimal segment of the rod, located between x and x + dx It has length dx and contains charge d = λ dx = cx dx Its distance from P is d + x and the potential it creates at P is dv d cx dx = d + x = d + x To find the total potential at P, we integrate over the length of the rod and obtain L c L xdx c c L V = = [ x dln( x+ d)] = L dln + πε d+ x πε πε d 9 m = (899 N m C )(89 C/m ) m (3 m)ln + 3 m 86 V =
Problem -35 We use E -: c V Ex ( x, y) = ( ) x ) y ( x = V/m 3V/m = V/m ) x; x V Ey ( x, y) = ( ) x ) y ( ) y = c V/m 3V/m h = 3V/m y y We evaluate at x = 3 m and y = m to obtain E = ( V/m)i ˆ+ ( V/m)j ˆ h Problem -53 (a) The potential energy is U = = ε d p c hc h 899 N m C 5 C m 9 6 = 5 J relative to the potential energy at infinite separation (b) Each sphere repels the other with a force that has magnitude = = ε d p 9 6 c899 N m C hc5 Ch m b g = 5 N ccording to Newton s second law the acceleration of each sphere is the force divided by the mass of the sphere Let m and m be the masses of the spheres The acceleration of sphere is 5 N a = = = 5 ms 3 m 5 kg and the acceleration of sphere is a 5 N = = = 5 ms 3 m kg (c) Energy is conserved The initial potential energy is U = 5 J, as calculated in part (a) The initial kinetic energy is zero since the spheres start from rest The final potential energy is zero since the spheres are then far apart The final kinetic energy is mv + mv, where v and v are the final velocities Thus, U = m v + m v
Momentum is also conserved, so = mv + mv These euations may be solved simultaneously for v and v Substituting v =( m/ m) v, from the momentum euation into the energy euation, and collecting terms, we obtain U = ( m / m)( m + m) v Thus, v 3 Um (5 J)( kg) 3 3 3 + + = = = 775 m/s m ( m m ) (5 kg)(5 kg kg) We thus obtain or v = 387 m/s v m = v = = m kg 3 5 kg 3 (775 m/s) 387 m/s, Problem -5 (a) Using U = V we can translate the graph of voltage into a potential energy graph (in ev units) rom the information in the problem, we can calculate its kinetic energy (which is its total energy at x = ) in those units: K i = 8 ev This is less than the height of the potential energy barrier (5 ev high once we ve translated the graph as indicated above) Thus, it must reach a turning point and then reverse its motion (b) Its final velocity, then, is in the negative x direction with a magnitude eual to that of its initial velocity That is, its speed (upon leaving this region) is 7 m/s Problem -67 (a) The magnitude of the electric field is 8 9 ( 3 C)( 899 N m C ) σ E = = = = ε ( ) πε 5 m (b) V = E = (5 m)( N/C) = 8 3 V (c) Let the distance be x Then which gives b g ΔV = V x V = I + x K J =5 V, ε p N C
V x = Δ V ΔV = b5 mgb5 Vg = 58 m 8V+ 5V Problem -99 (a) The charge on every part of the ring is the same distance from any point P on the axis This distance is r = z +, where is the radius of the ring and z is the distance from the center of the ring to P The electric potential at P is V = d d d πε = r πε = z πε = + z + πε z + (b) The electric field is along the axis and its component is given by V / 3/ ( ) ( ) ( ) z E = = z + = z + z = z πε z πε πε ( z + ) 3/ This agrees with E 3-6