Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 16 CHAPTER 16

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1 CHAPTER The number of electrons is N = Q/e = ( C)/( C/electrons) = electrons.. The mgnitude of the Coulomb force is Q /r. If we divide the epressions for the two forces, we hve F / = (r 1 /r ) ; F /( N) = (8), which gives F =.7 N. 3. The mgnitude of the Coulomb force is Q /r. If we divide the epressions for the two forces, we hve F / = (r 1 /r ) ; 3 = [(0.0 cm)/r ], which gives r = 11.5 cm. 4. The mgnitude of the Coulomb force is Q /r. If we divide the epressions for the two forces, we hve F / = (r 1 /r ) ; F /(0.000 N) = (150 cm/30.0 cm), which gives F = N. 5. The mgnitude of the Coulomb force is Q /r = ( N á m /C )(6)( C)( C)/( m) = N. 6. The mgnitude of the Coulomb force is Q /r = ( N á m /C )( C)( C)/( m) = 9. N. 7. The mgnitude of the Coulomb force is Q /r = ( N á m /C )( C)( C)/(0.40 m) = N. 8. The number of ecess electrons is N = Q/e = ( C)/( C/electrons) = electrons. The mss increse is m = Nm e = ( electrons)( kg/electron) = kg. 9. Becuse the chrge on the Erth cn be considered to be t the center, we cn use the epression for the force between two point chrges. For the Coulomb force to be equl to the weight, we hve kq /R = mg; Pge 16 1

2 ( N á m /C )Q /( m) = (1050 kg)(9.80 m/s ), which gives Q = C. Pge 16

3 10. The number of molecules in 1.0 kg is N = [(1.0 kg)(10 3 g/kg)/(18 g/mol)]( molecules/mol) = molecules. Ech molecule of H O contins (1) 8 = 10 electrons. The chrge of the electrons in 1.0 kg is q = ( molecules)(10 electrons/molecule)( C/electron) = C. 11. Using the smbols in the figure, we find the mgnitudes of the three individul forces: = = kq 1 Q /r 1 = kq 1 Q / = ( N á m /C )( C)( C)/(0.35 m) = N. 3 = F 31 = kq 1 Q 3 /r 1 = kq 1 Q 3 /() = ( N á m /C )( C)( C)/[(0.35 m)] = N. F 3 = F 3 = kq Q 3 /r 1 = kq Q 3 / = ( N á m /C )( C)( C)/(0.35 m) =.8 10 N. The directions of the forces re determined from the signs of the chrges nd re indicted on the digrm. For the net forces, we get = 3 = N N = N (left). F = F 3 = N.8 10 N = F 3 = F 31 F 3 = N.8 10 N = Note tht the sum for the three chrges is zero N (right) N (left). Q 1 Q Q 3 3 F 31 F 3 F 3 1. Becuse ll the chrges nd their seprtions re equl, we find the mgnitude of the individul forces: = kqq/ = kq / = ( N á m /C )( C) /(0.150 m) Q = 48.4 N. The directions of the forces re determined from the signs of the chrges nd re indicted on the digrm. For the forces on the top chrge, we see tht the horizontl Q 60 Q components will cncel. For the net force, we hve F = cos 30 cos 30 = cos 30 = (48.4 N) cos 30 = 83.8 N up, or w from the center of the tringle. From the smmetr ech of the other forces will hve the sme mgnitude nd direction w from the center: The net force on ech chrge is 83.8 N w from the center of the tringle. Note tht the sum for the three chrges is zero. Pge 16 3

4 13. We find the mgnitudes of the individul forces on the chrge t the upper right corner: = F = kqq/ = kq / = ( N á m /C )( C) /(1.00 m) = N. F 3 = kqq/(ì) = kq / = ( N á m /C )( C) /(1.00 m) = N. The directions of the forces re determined from the signs of the chrges nd re indicted on the digrm. Q Q For the forces on the top chrge, we see tht the net force will be long the digonl. For the net force, we hve F = cos 45 F cos 45 F 3 = ( N) cos N = N long the digonl, or w from the center of the squre. From the smmetr, ech of the other forces will hve the sme mgnitude nd direction w from the center: The net force on ech chrge is N w from the center of the squre. Note tht the sum for the three chrges is zero. Q Q F 3 F 14. Becuse the mgnitudes of the chrges nd the distnces hve not chnged, we hve the sme mgnitudes of the individul forces on the chrge t the upper right corner: = F = kqq/ = N. F 3 = kq / = N. The directions of the forces re determined from the signs of the chrges nd re indicted on the digrm. For the forces on the top chrge, we see tht the net force will be long the digonl. For the net force, we hve F F = cos 45 F cos 45 F 3 Q Q = ( N) cos N = N long the digonl, or towrd the center of the squre. From the smmetr, ech of the other forces will hve the sme mgnitude nd direction towrd the center: The net force on ech chrge is N towrd the center of the squre. Note tht the sum for the three chrges is zero. Q Q F For the two forces, we hve F electric = kq 1 q /r 1 = ke /r = ( N á m /C )( C) /( m) = N. F grvittionl = Gm 1 m /r = ( N á m /kg )( kg)( kg)/( m) = N. The rtio of the forces is F electric /F grvittionl = ( N)/( N) = Pge 16 4

5 16. Becuse the electricl ttrction must provide the sme force s the grvittionl ttrction, we equte the two forces: kqq/r = Gm 1 m /r ; ( N á m /C )Q = ( N á m /kg )( kg)( kg), which gives Q = C. 17. If the seprtion is r nd one chrge is Q 1, the other chrge will be Q = Q T Q 1. For the repulsive force, we hve Q /r = kq 1 (Q T Q 1 )/r F. () If we plot the force s function of Q 1, we see tht the mimum occurs when Q 1 =!Q T, which we would epect from smmetr, since we could interchnge the two chrges without chnging the force. 0 (b) We see from the plot tht the minimum occurs when either chrge is zero: Q 1 (or Q ) = 0. Q T Q 18. The ttrctive Coulomb force provides the centripetl ccelertion of the electron: ke /r = mv /r, or r = ke /mv ; r = ( N á m /C )( C) /( kg)( m/s) = m. 19. If we plce positive chrge, it will be repelled b the positive chrge nd ttrcted b the negtive chrge. Thus the third chrge must be plced long the line of the chrges, but not between them. For the net force to be zero, the mgnitudes of the individul forces must be equl: Q/r 1 = kq Q/r, or Q 1 /( ) = Q / ; (5.7 µc)/(0.5 m ) = (3.5 µc)/, which gives = 0.91 m, 0.11 m. The negtive result corresponds to the position between the chrges where the mgnitudes nd the directions re the sme. Thus the third chrge should be plced 0.91 m beond the negtive chrge. Note tht we would hve the sme nlsis if we used negtive chrge. Q F F 1 Q Q P 0. If one chrge is Q 1, the other chrge will be Q = Q Q 1. For the force to be repulsive, the two chrges must hve the sme sign. Becuse the totl chrge is positive, ech chrge will be positive. We ccount for this b considering the force to be positive: Q /r = kq 1 (Q Q 1 )/r ; 1.0 N = ( N á m /C )Q 1 ( C Q 1 )/(1.06 m), which is qudrtic eqution: Q 1 ( C)Q C = 0, which gives Q 1 = C, C. Note tht, becuse the lbels re rbitrr, we get the vlue of both chrges. For n ttrctive force, the chrges must hve opposite signs, so their product will be negtive. We ccount for this b considering the force to be negtive: Q /r = kq 1 (Q Q 1 )/r ; 1.0 N = ( N á m /C )Q 1 ( C Q 1 )/(1.06 m), which is qudrtic eqution: Q 1 ( C)Q C = 0, which gives Q 1 = C, C. Pge 16 5

6 1. The ccelertion is produced b the force from the electric field: F = qe = m; ( C)(600 N/C) = ( kg), which gives = m/s. Becuse the chrge on the electron is negtive, the direction of force, nd thus the ccelertion, is opposite to the direction of the electric field. The direction of the ccelertion is independent of the velocit.. If we tke the positive direction to the est, we hve F = qe = ( C)( 3500 N/C) = N, or N (west). 3. If we tke the positive direction to the south, we hve F = qe ; N = ( C)E, which gives E = N/C (south). 4. If we tke the positive direction up, we hve F = qe ; 8.4 N = ( C)E, which gives E = N/C (up). 5. The electric field bove positive chrge will be w from the chrge, or up. We find the mgnitude from E = kq/r = ( N á m /C )( C)/(0.300 m) = N/C (up). 6. The directions of the fields re determined from the signs of the chrges nd re indicted on the digrm. The net electric field will be to the left. We find its mgnitude from E = kq 1 / kq / = k(q 1 Q )/ = ( N á m /C )( C C)/(0.00 m) = N/C. Thus the electric field is N/C towrd the negtive chrge. E Q 1 Q 7. The ccelertion is produced b the force from the electric field: F = qe = m; ( C)E = ( kg)(15 m/s ), which gives E = N/C. Becuse the chrge on the electron is negtive, the direction of force, nd thus the ccelertion, is opposite to the direction of the electric field, so the electric field is N/C (south). 8. The directions of the fields re determined from the signs of the chrges nd re in the sme direction, s indicted on the digrm. The net electric field will be to the left. We find its mgnitude from E = kq 1 / kq / = k(q Q)/ = kq/ 1750 N/C = ( N á m /C )Q/(0.080 m), which gives Q = C. E Q Q Pge 16 6

7 9. From the digrm, we see tht the electric fields produced b the chrges will hve the sme mgnitude, nd the resultnt field will be down. The distnce from the origin is, so we hve E = E = kq/r = kq/( ). From the smmetr, for the mgnitude of the electric field we hve E = E sin θ = [kq/( )][/( ) 1/ ] = kq/( ) 3/ prllel to the line of the chrges. Q Q r θ E E E 30. At point A, from the digrm, we see tht the electric fields produced b the chrges will hve the sme mgnitude, nd the resultnt field will be up. We find the ngle θ from E A E tn θ = (0.050 m)/(0.100 m) = 0.500, or θ = 6.6. A A For the mgnitudes of the individul fields we hve θ A (0, 5) A = E A = kq/r A Q = ( N á m /C )( C)/[(0.100 m) (0.050 m) θ ] Q = N/C. (10, 0) (10, 0) From the smmetr, the resultnt electric field is E A = A sin θ = ( N/C) sin 6.6 = N/C up. For point B we find the ngles for the directions of the fields from E B tn θ 1 = (0.050 m)/(0.050 m) = 1.00, or θ 1 = tn θ = (0.050 m)/(0.150 m) = 0.333, or θ = EB B For the mgnitudes of the individul fields we hve B(5, 5) Q B = kq/r θ 1 θ 1B Q = ( N á m /C )( C)/[(0.050 m) (0.050 m) ] = N/C. E B = kq/r B = ( N á m /C )( C)/[(0.150 m) (0.050 m) ] = N/C. For the components of the resultnt field we hve (10, 0) (10, 0) E B = B cos θ 1 E B cos θ = ( N/C) cos 45.0 ( N/C) cos 18.4 = N/C; E B = B sin θ 1 E B sin θ = ( N/C) sin 45.0 ( N/C) sin 18.4 = N/C. We find the direction from tn θ B = E B /E B = ( N/C)/( N/C) = 1.49, or θ 1 = 56.. We find the mgnitude from E B = E B /cos θ B = ( N/C)/cos 56. = N/C. Thus the field t point B is N/C 56 bove the horizontl. Pge 16 7

8 31. The directions of the individul fields will be long the digonls of the squre, s shown. We find the mgnitudes of the individul fields: = kq 1 /(/Ì) = kq 1 / = ( N á m /C )( C)/(0.60 m) = N/C. E = E 3 = E 4 = kq /(/Ì) = kq / = ( N á m /C )( C)/(0.60 m) = N/C. From the smmetr, we see tht the resultnt field will be long the digonl shown s the -is. For the net field, we hve E = E 3 = N/C N/C = N/C. Thus the field t the center is N/C w from the positive chrge. Q Q E4 E 3 E E Q 1 Q 3. The directions of the individul fields re shown in the figure. We find the mgnitudes of the individul fields: = E 3 = kq/ = ( N á m /C )( C)/(1.00 m) = N/C. E = kq/(ì) =!kq / =!( N á m /C )( C)/(1.00 m) = N/C. From the smmetr, we see tht the resultnt field will be long the digonl shown s the -is. For the net field, we hve E = cos 45 E = ( N/C) cos N/C = N/C. Thus the field t the unoccupied corner is N/C w from the opposite corner. E 3 Q Q Q E Pge 16 8

9 33. () The directions of the individul fields re shown in the figure. We find the mgnitudes of the individul fields: = E = kq/. For the components of the resultnt field we hve E = E sin 60 = 0.866kQ/ ; E = E cos 60 = kq/ 0.500kQ/ = 1.50kQ/. We find the direction from tn θ = E /E = ( 1.50kQ/ )/( 0.866kQ/ ) = 1.73, or θ = 60. We find the mgnitude from E = E /cos θ = (0.866kQ/ )/cos 60 = 1.73kQ/. Thus the field is 1.73kQ/ 60 below the -is. (b) The directions of the individul fields re shown in the figure. The mgnitudes of the individul fields will be the sme: = E = kq/. For the components of the resultnt field we hve E = E sin 60 = 0.866kQ/ ; E = E cos 60 = kq/ 0.500kQ/ = 0.500kQ/. We find the direction from tn θ = E /E = ( 0.500kQ/ )/( 0.866kQ/ ) = 0.577, or θ = 30. We find the mgnitude from E = E /cos θ = (0.866kQ/ )/cos 30 = kq/. Thus the field is kq/ 30 below the -is. O E O Q 60 Q 60 E Q Q 34. q 3q 35. The ccelertion is produced b the force from the electric field: F = qe = m; ( C)E = ( kg)( )(9.80 m/s ), which gives E = 0.10 N/C. 36. If we let be the distnce from the center of the Erth, we hve GM Moon /(D ) = GM Erth / = 81GM Moon /, or 81(D ) =. When we tke the squre root of both sides, we get = 9D/10 = 9( km)/10 = km from the center of the Erth. Note tht tking negtive squre root gives = 9D/8, the point on the other side of the Moon where the mgnitudes re equl, but the fields hve the sme direction. Pge 16 9

10 37. For the electric field to be zero, the individul fields must hve opposite directions, so the two chrges must hve the sme sign. For the net field to be zero, the mgnitudes of the individul fields must be equl: E = kq 1 /r 1 = kq /r, or Q 1 /(@) = Q /(%), which gives Q = 4Q 1. /3 Q 1 E Q 38. () We find the ccelertion produced b the electric field: F = qe = m; ( C)( N/C) = ( kg), which gives = m/s. Becuse the field is constnt, the ccelertion is constnt, so we find the speed from v = v 0 = 0 ( m/s )(0.010 m), which gives v = m/s. (b) For the rtio of the two forces, we hve mg/qe = ( kg)(9.80 m/s )/( C)( N/C) = Thus mg Ç qe. 39. () The ccelertion of the electron, nd thus the force produced b the electric field, must be opposite its velocit. Becuse the electron hs negtive chrge, the direction of the electric field will be opposite tht of the force, so the direction of the electric field is in the direction of the velocit, to the right. (b) Becuse the field is constnt, the ccelertion is constnt, so we find the required ccelertion from v = v 0 ; 0 = [0.01( m/s)] (0.050 m), which gives = m/s. We find the electric field from F = qe = m; ( C)E = ( kg)( m/s ), which gives E = N/C. Pge 16 10

11 40. () To estimte the force between thmine nd n denine, we ssume tht onl the toms with n indicted chrge mke contribution. Becuse ll chrges re frctions of the electronic chrge, we let Q H = Q N = f 1 e, nd Q O = Q C = f e. A convenient numericl fctor will be ke /(10 10 m/ ) = ( N á m /C )( C) /(10 10 m/ ) = N á. For the first contribution we find the force for the bond of the ogen on the thmine with the H-N pir on the denine. From Newton's third lw, we know tht the force on one must equl the force on the other. We find the ttrctive force on the ogen: F O = kq O {[Q H /( 1 ) ] (Q N / 1 )} = ke f f 1 {[1/( 1 ) ] (1/ 1 )} = ( N á )(0.4)(0.){[1/( ) ] [1/(.80 ) ]} = N. For the force for the lower bond of the H-N pir on the thmine with the nitrogen on the denine, we find the ttrctive force on the nitrogen: F N = kq N {[Q H /( ) ] (Q N / )} = ke f 1 f 1 {[1/( ) ] (1/ )} = ( N á )(0.)(0.){[1/( ) ] [1/(3.00 ) ]} = N. There will be repulsive force between the ogen of the first bond nd the nitrogen of the second bond. To find the seprtion of the two, we note tht the distnce between the two nitrogens of the denine, which is pproimtel perpendiculr to 1, is cos 30 = We find the mgnitude of this force from F O-N = kq O {Q N /[ 1 (1.73) ]} = ke f f 1 {1/[ 1 (1.73) ]} = ( N á )(0.4)(0.){1/[(.80 ) (1.73 ) ]} = N. We find the ngle tht this force mkes with the line of the other bonds from tn θ = 1.73/ 1 = 1.73 /.89 = 0.6, or θ = 3. Thus the component tht contributes to the bond is ( N) cos 3 = N. The other contribution will be from the crbon tom on the thmine. Becuse the distnce is slightl greter nd there will be ttrction to the nitrogens nd repulsion from the hdrogen, we neglect this contribution. Thus the estimted net bond is N N N N. (b) To estimte the net force between ctosine nd gunine, we note tht there re two ogen bonds, one nitrogen bond, nd one repulsive O-N force. We neglect the other forces becuse the involve cncelltion from the involvement of both hdrogen nd nitrogen. If we ignore the smll chnge in distnces, we hve ( N) N N N. (c) The totl force for the DNA molecule is ( N N)(10 5 pirs) 10 4 N. C C T 10 O N O H 1 H H N N C C H A 41. When we equte the two forces, we hve mg = ke /r ; ( kg)(9.80 m/s ) = ( N á m /C )( C) /r, which gives r = 5.08 m. 4. Becuse copper tom hs 9 electrons, we find the number of electrons in the penn from N = [(3.0 g)/(63.5 g/mol)]( toms/mol)(9 electrons/tom) = electrons. We find the frctionl loss from q/q = ( C)/( electrons)( C/electron) = Pge 16 11

12 Pge 16 1

13 43. The weight must be blnced b the force from the electric field: mg = qe; ( kg)(9.80 m/s ) = ( C)E, which gives E = N/C (up). 44. Becuse we cn tret the chrge on the Erth s point chrge t the center, we hve E = kq/r ; 150 N/C = ( N á m /C )Q/( m), which gives Q = C. Becuse the field points towrd the center, the chrge must be negtive. 45. The weight must be blnced b the force from the electric field: mg = ρ9?r 3 g = NeE ; (1000 kg/m 3 )9?( m) 3 (9.80 m/s ) = N( C)(150 N/C), which gives N = electrons. 46. The directions of the individul fields will be long the digonls of the squre, s shown. All distnces re the sme. We find the mgnitudes of the individul fields: = kq 1 /(/Ì) = kq 1 / = ( N á m /C )( C)/(0.5 m) = N/C. E = kq /(/Ì) = = ( N/C) = N/C. E 3 = kq 3 /(/Ì) = 3 = 3( N/C) = N/C. E 4 = kq 4 /(/Ì) = 4 = 4( N/C) = N/C. We simplif the vector ddition b using the -coordinte sstem shown. For the components of the resultnt field we hve E = E 4 E = N/C N/C = N/C; E = E 3 = N/C N/C = N/C. Thus we see tht the resultnt will be in the -direction: E = E cos 45 = ( N/C) cos 45 = N/C up. Q 1 Q 4 E E 3 E 4 Q Q We find the force between the groups b finding the force on the CO group from the HN group. A convenient numericl fctor will be ke /(10 9 m/nm) = ( N á m /C )( C) /(10 9 m/nm) = N á nm. For the forces on the toms, we hve F O = kq O {[Q H /( d ) ] (Q N / )} = ke f O f H {[1/( d ) ] (1/ )} = ( N á nm )(0.4)(0.){[1/(0.8 nm 0.10 nm) ] [1/(0.8 nm) ]} = N. F C = kq C {[Q N /( d 1 ) ] Q H /( d 1 d ) ]} = ke f C f N {[1/( d 1 ) ] [1/( d 1 d ) ]} = ( N á nm )(0.4)(0.){[1/(0.8 nm 0.1 nm) ] [1/(0.8 nm 0.1 nm 0.10 nm) ]} = N. Thus the net force is F = F O F C = N N = N (ttrction). C O H N d 1 d Pge 16 13

14 48. Becuse the chrges hve the sme sign, the repel ech other. The force from the third chrge must blnce the repulsive force for ech chrge, so the third chrge must be positive nd between the two negtive chrges. For ech of the negtive chrges, we hve Q 0 : kq 0 Q/ = kq 0 (3Q 0 )/, or Q = 3 Q 0 ; Q 0 F F Q 3Q 0 3Q 0 : k3q 0 Q/( ) = kq 0 (3Q 0 )/, or Q = ( ) Q 0. Thus we hve 3 = ( ), which gives = 1.37, Becuse the positive chrge must be between the chrges, it must be from Q 0. When we use this vlue in one of the force equtions, we get Q = 3(0.366 ) Q 0 / = 0.40Q 0. Thus we plce chrge of 0.40Q from Q 0. Note tht the force on the middle chrge is lso zero. 49. Becuse the chrge moves in the direction of the electric field, it must be positive. We find the ngle of the string from the dimensions: cos θ = (0.49 m)/(0.50 m) = 0.98, or θ = Becuse the chrge is in equilibrium, the resultnt force is zero. We see from the force digrm tht tn θ = QE/mg; tn 11.5 = Q(900 N/C)/( kg)(9.80 m/s ), which gives Q = C. θ E F T Q F T mg θ QE QE mg 50. Becuse the chrges hve opposite signs, the loction where the electric field is zero must be outside the two chrges, s shown. The fields from the two chrges must blnce: P kq 1 / = kq /( ) ; Q 1 Q E ( C)/ = ( C)/(.0 m ), which gives = 1.4 m, 3.6 m. Becuse 1.4 m is between the chrges, the loction is 3.6 m from the positive chrge, nd 1.6 m from the negtive chrge. 51. () The force is opposite to the direction of the electron. We find the ccelertion produced b the electric field: qe = m; ( C)( N/C) = ( kg), which gives = m/s. Becuse the field is constnt, the ccelertion is constnt, so we find the distnce from v = v 0 ; 0 = ( m/s) ( m/s ), which gives = m = 0.83 mm. (b) We find the time from = v 0 t!t ; 0 = ( m/s)t!( m/s )t, which gives t = 0 (the strting time), nd s =. ns. Pge 16 14

15 5. The ngulr frequenc of the SHM is ω = (k/m) 1/ = [(16 N/m)/(0.800 kg)] 1/ = 1.5 s 1. If we tke down s positive, with respect to the equilibrium position, the bll will strt t mimum displcement, so the position s function of time is = A cos(ωt) = ( m) cos [(1.5 s 1 )t]. Becuse the chrge is negtive, the electric field t the tble will be up nd the distnce from the tble is r = H = m ( m) cos [(1.5 s 1 )t]. The electric field is E = kq/r = ( N á m /C )( C)/{0.150 m ( m) cos [(1.5 s 1 )t]} = ( N/C)/{3 cos [(1.5 s 1 )t]} up. 53. We consider the forces on one bll. (The other will be the sme ecept for the reversl.) The seprtion of the chrges is r = sin 30 = (0.70 m) sin 30 = 0.70 m. From the equilibrium force digrm, we hve tn θ = F/mg = [k(!q)(!q)/r ]/mg; tn 30 = (( N á m /C )Q / (0.70 m) ( kg)(9.80 m/s ), which gives Q = C = 5.4 µc. Q/ θ F T F T mg θ F Q/ F mg 54. The pe will dischrge when the electric field t the surfce eceeds the brekdown field. Becuse we cn tret the chrge on the pe s point chrge t the center, we hve E = kq/r ; N/C = ( N á m /C )Q/( m), which gives Q = C. 55. We find the electric field t the loction of Q 1 due to the pltes nd Q. For the field of Q we hve E = kq / = ( N á m /C )( C)/(0.34 m) = N/C (left). The field from the pltes is to the right, so we hve E net = E pltes E = 73,000 N/C N/C = N/C (left). For the force on Q 1, we hve = Q 1 E net = ( C)( N/C) = 0.19 N (right). E Q 1 E pltes Q 56. We tke up s the positive direction nd ssume tht E is up. From the equilibrium force digrm, we hve F T QE = mg; 5.67 N ( C)E = (0.10 kg)(9.80 m/s ), which gives E = N/C (down). QE F T Q mg Pge 16 15

16 Pge 16 16