Page 1 8. As ball are in equi lib rium F T e = sin a mg T = cos a F mg e = tan a q r 2 0 2 4 = pe a tan Here, r l =2 sina q l 2 0 2 2 16 = pe a a sin tan q = ´ - 3.3 10 8 C. 9. Same as Q.7. In tro duc tory Ex er cise 21.3. 10. See Q.7. In tro duc tory Ex er cise 21.3. 11. E q r = × ® 1 4 1 0 3 pe r = ´ ´ - ´ + - - 9 10 10 9 9 2 232 ( ) (( ) ( )) ( ) / ^ ^ 8.0 1.2 1.6 1.2 1.6 i j =- - 18 2( ) ^ ^ 1.2 1.6 i j N/C. 12. Con sider an el e men tary por tion on the ring of length dl sub t end ing an gle df at cen tre ‘O’ of the ring. Charge on this portion, dq dl Rd = = f l l \ dE dq R d R = × = f 1 4 1 4 0 2 0 pe pe l Here, dE sin f components of field will cancel each other. Hence, Net field at O E dE R d = f = × f f ò ò - cos cos / / 1 4 0 2 2 pe l p p = × 1 4 2 0 pe l R 13. Con sider el e men tary por tion of the rod of length dl at a dis tance l from the cen tre O of the rod. Charge on this portion dq dl Q L dl = = l \ dE dq a = × f 1 4 0 2 pe ( ) sec = × f 1 4 0 2 2 pe Q dl La sec Now, l a = f tan Þ dl a d = f f sec 2 \ dE Qd La = × f 1 4 0 pe Net Electric field at P. E dE = f ò cos [ dEsin f components will cancel each other as rod in symmetrical about P.] = × f f - ò 1 4 0 pe q q Q La d cos = × 1 4 2 0 pe q Q La sin But sinq= + æ è ç ö ø ÷ = + L a L L a L 2 2 4 2 2 2 2 \ E Q a a L = × + 1 4 2 4 0 2 2 pe 38 dE dE cos f dE cos f dE dE sin f dE sin f O f R df dl dl dE dE cos f dE cos f dE dE sin f dE sin f df Q P dl l dl f O a O a F e F e mg T sin a mg r T cos a T T Page 2 8. As ball are in equi lib rium F T e = sin a mg T = cos a F mg e = tan a q r 2 0 2 4 = pe a tan Here, r l =2 sina q l 2 0 2 2 16 = pe a a sin tan q = ´ - 3.3 10 8 C. 9. Same as Q.7. In tro duc tory Ex er cise 21.3. 10. See Q.7. In tro duc tory Ex er cise 21.3. 11. E q r = × ® 1 4 1 0 3 pe r = ´ ´ - ´ + - - 9 10 10 9 9 2 232 ( ) (( ) ( )) ( ) / ^ ^ 8.0 1.2 1.6 1.2 1.6 i j =- - 18 2( ) ^ ^ 1.2 1.6 i j N/C. 12. Con sider an el e men tary por tion on the ring of length dl sub t end ing an gle df at cen tre ‘O’ of the ring. Charge on this portion, dq dl Rd = = f l l \ dE dq R d R = × = f 1 4 1 4 0 2 0 pe pe l Here, dE sin f components of field will cancel each other. Hence, Net field at O E dE R d = f = × f f ò ò - cos cos / / 1 4 0 2 2 pe l p p = × 1 4 2 0 pe l R 13. Con sider el e men tary por tion of the rod of length dl at a dis tance l from the cen tre O of the rod. Charge on this portion dq dl Q L dl = = l \ dE dq a = × f 1 4 0 2 pe ( ) sec = × f 1 4 0 2 2 pe Q dl La sec Now, l a = f tan Þ dl a d = f f sec 2 \ dE Qd La = × f 1 4 0 pe Net Electric field at P. E dE = f ò cos [ dEsin f components will cancel each other as rod in symmetrical about P.] = × f f - ò 1 4 0 pe q q Q La d cos = × 1 4 2 0 pe q Q La sin But sinq= + æ è ç ö ø ÷ = + L a L L a L 2 2 4 2 2 2 2 \ E Q a a L = × + 1 4 2 4 0 2 2 pe 38 dE dE cos f dE cos f dE dE sin f dE sin f O f R df dl dl dE dE cos f dE cos f dE dE sin f dE sin f df Q P dl l dl f O a O a F e F e mg T sin a mg r T cos a T T 14. (a) As shown in fig ure, di rec tion of elec tric field at P will be along + ve y-axis. (b) Positive x-axis. (c) Positive y-axis. 15. Let E q R 1 0 2 1 4 = × pe Resultant fields of two opposite charges can be shown as given in figure. Clearly resultant field is along angle bisector of field towards 9 and 10. Hence time shown by clock in the direction of electric field is 9 : 30. 16. (a) a F m eE m = = - = - ´ ´ ´ ´ - - 1.6 9.1 10 1 10 10 19 3 31 = - ´ 1.76 10 14 ms -2 u = ´ 5.00 10 8 cm/s = ´ 5 10 6 ms -1 v = 0 v u as 2 2 2 - = s = ´ ´ ´ = ´ = - ( ) 5 10 2 10 1 10 6 2 14 2 1.7 .4 1.4 cm (b) v u at = + t = ´ ´ = ´ = - 5 10 10 10 28 6 14 8 1.76 2.8 ns. (c) Dk = work done by electric field. = × = - F x eEx = - ´ ´ ´ ´ ´ - - 1.6 10 1 10 8 10 19 3 3 = - ´ - 1.28 J 10 18 Loss of KE = ´ - 1.28 J 10 18 17. Here, u u x = ° = cos45 25 2 ms -1 u u y = ° = sin45 25 2 ms -1 a qE x = = ´ ´ ´ - 2 10 2 10 6 7 =40 ms -1 a y = -10 ms -1 y u a t yt y = + 1 2 y t t = - 25 2 5 2 39 E 1 E E 2 P x y –Q Q y +Q +Q E 2 E 1 E x P –Q +Q E 1 E 2 E P 12 E 12 2 3 4 5 6 7 8 9 10 11 E 1 E 2 E 3 E 4 E 5 E 6 E 7 E 8 E 9 E 10 E 11 1 6E 1 6E 1 6E 1 6E 1 E 6E 1 6E 1 q u E Page 3 8. As ball are in equi lib rium F T e = sin a mg T = cos a F mg e = tan a q r 2 0 2 4 = pe a tan Here, r l =2 sina q l 2 0 2 2 16 = pe a a sin tan q = ´ - 3.3 10 8 C. 9. Same as Q.7. In tro duc tory Ex er cise 21.3. 10. See Q.7. In tro duc tory Ex er cise 21.3. 11. E q r = × ® 1 4 1 0 3 pe r = ´ ´ - ´ + - - 9 10 10 9 9 2 232 ( ) (( ) ( )) ( ) / ^ ^ 8.0 1.2 1.6 1.2 1.6 i j =- - 18 2( ) ^ ^ 1.2 1.6 i j N/C. 12. Con sider an el e men tary por tion on the ring of length dl sub t end ing an gle df at cen tre ‘O’ of the ring. Charge on this portion, dq dl Rd = = f l l \ dE dq R d R = × = f 1 4 1 4 0 2 0 pe pe l Here, dE sin f components of field will cancel each other. Hence, Net field at O E dE R d = f = × f f ò ò - cos cos / / 1 4 0 2 2 pe l p p = × 1 4 2 0 pe l R 13. Con sider el e men tary por tion of the rod of length dl at a dis tance l from the cen tre O of the rod. Charge on this portion dq dl Q L dl = = l \ dE dq a = × f 1 4 0 2 pe ( ) sec = × f 1 4 0 2 2 pe Q dl La sec Now, l a = f tan Þ dl a d = f f sec 2 \ dE Qd La = × f 1 4 0 pe Net Electric field at P. E dE = f ò cos [ dEsin f components will cancel each other as rod in symmetrical about P.] = × f f - ò 1 4 0 pe q q Q La d cos = × 1 4 2 0 pe q Q La sin But sinq= + æ è ç ö ø ÷ = + L a L L a L 2 2 4 2 2 2 2 \ E Q a a L = × + 1 4 2 4 0 2 2 pe 38 dE dE cos f dE cos f dE dE sin f dE sin f O f R df dl dl dE dE cos f dE cos f dE dE sin f dE sin f df Q P dl l dl f O a O a F e F e mg T sin a mg r T cos a T T 14. (a) As shown in fig ure, di rec tion of elec tric field at P will be along + ve y-axis. (b) Positive x-axis. (c) Positive y-axis. 15. Let E q R 1 0 2 1 4 = × pe Resultant fields of two opposite charges can be shown as given in figure. Clearly resultant field is along angle bisector of field towards 9 and 10. Hence time shown by clock in the direction of electric field is 9 : 30. 16. (a) a F m eE m = = - = - ´ ´ ´ ´ - - 1.6 9.1 10 1 10 10 19 3 31 = - ´ 1.76 10 14 ms -2 u = ´ 5.00 10 8 cm/s = ´ 5 10 6 ms -1 v = 0 v u as 2 2 2 - = s = ´ ´ ´ = ´ = - ( ) 5 10 2 10 1 10 6 2 14 2 1.7 .4 1.4 cm (b) v u at = + t = ´ ´ = ´ = - 5 10 10 10 28 6 14 8 1.76 2.8 ns. (c) Dk = work done by electric field. = × = - F x eEx = - ´ ´ ´ ´ ´ - - 1.6 10 1 10 8 10 19 3 3 = - ´ - 1.28 J 10 18 Loss of KE = ´ - 1.28 J 10 18 17. Here, u u x = ° = cos45 25 2 ms -1 u u y = ° = sin45 25 2 ms -1 a qE x = = ´ ´ ´ - 2 10 2 10 6 7 =40 ms -1 a y = -10 ms -1 y u a t yt y = + 1 2 y t t = - 25 2 5 2 39 E 1 E E 2 P x y –Q Q y +Q +Q E 2 E 1 E x P –Q +Q E 1 E 2 E P 12 E 12 2 3 4 5 6 7 8 9 10 11 E 1 E 2 E 3 E 4 E 5 E 6 E 7 E 8 E 9 E 10 E 11 1 6E 1 6E 1 6E 1 6E 1 E 6E 1 6E 1 q u E at the end of motion, t T = and y = 0 \ T = 5 2 s Also at the end of motion, x R = \ x u t a t x x = + 1 2 2 R= ´ + ´ æ è ç ö ø ÷ 25 2 5 2 20 5 2 2 =312.5 m 18. (a) R qE = m q 2 2 sin sin2 2 q= qER mu = ´ ´ ´ ´ ´ ´ ´ - - - 1.6 1.27 1.67 9.55 10 720 10 10 10 19 3 27 3 2 ( ) =0.96 2 88 q= ° or 92° q= ° 44 or 46° T mh E = 2 2 sinq = ´ ´ ´ ´ ´ ´ ´ - - 2 10 1 2 10 10 720 3 31 19 9.55 1.67 1.6 = ´ - 1.95 10 11 s 19. (a) a E j ® ® - - = - = - ´ ´ ´ e m 1.6 9.1 10 120 10 19 31 ^ = - ´ 2.1 10 13 i ^ m/s (b) t x u x = = ´ ´ = ´ - - D 2 10 10 4 3 10 2 5 7 1.5 s v u a t y y y = + = ´ ´ ´ ´ ´ - 3.0 2.1 10 10 4 3 10 6 13 7 = ´ 0.2 10 6 m/s v i j ® = ´ + ´ ( ) ( ) ^ ^ 1.5 0.2 10 10 5 6 20. Ab so lute po ten tial can be zero at two points on the x-axis. One in be tween the charges and other on the left of charge a 1 (smaller in mag ni tu de). Case I. In between two charges : let potential is zero at a distance x from q 1 towards q 2 . V q x q x = × + × - = 1 4 1 4 100 0 0 1 0 2 pe pe = × ´ - × ´ - = - - 1 4 2 10 1 4 3 10 100 0 0 6 0 6 pe pe x x Þ 200 2 3 - = x x x =20 cm Case II. Consider the potential is zero at a distance x from charge q, on its left. \ V q x q x = × + × + = 1 4 1 4 100 0 0 1 0 2 pe pe = × ´ + × - ´ + = - - 1 4 2 10 1 4 3 10 100 0 0 6 0 6 pe pe x x 200 2 3 + = x x x =200 cm 21. Let us first find the po ten tial at a point on the per pen dic u lar bi sec tor of a line charge. Consider a line of carrying a line charge density l having length L. Consider an elementary portion of length dl on the rod. Charge on this portion dq dl = l \ dV dl r = × f 1 4 0 pe l sec Now, l r = f tan dl r d = f f sec 2 40 O 100cm q 1 q 2 X q 1 q 2 x 100–x q 1 q 2 x 100 cm f q L r l dl Page 4 8. As ball are in equi lib rium F T e = sin a mg T = cos a F mg e = tan a q r 2 0 2 4 = pe a tan Here, r l =2 sina q l 2 0 2 2 16 = pe a a sin tan q = ´ - 3.3 10 8 C. 9. Same as Q.7. In tro duc tory Ex er cise 21.3. 10. See Q.7. In tro duc tory Ex er cise 21.3. 11. E q r = × ® 1 4 1 0 3 pe r = ´ ´ - ´ + - - 9 10 10 9 9 2 232 ( ) (( ) ( )) ( ) / ^ ^ 8.0 1.2 1.6 1.2 1.6 i j =- - 18 2( ) ^ ^ 1.2 1.6 i j N/C. 12. Con sider an el e men tary por tion on the ring of length dl sub t end ing an gle df at cen tre ‘O’ of the ring. Charge on this portion, dq dl Rd = = f l l \ dE dq R d R = × = f 1 4 1 4 0 2 0 pe pe l Here, dE sin f components of field will cancel each other. Hence, Net field at O E dE R d = f = × f f ò ò - cos cos / / 1 4 0 2 2 pe l p p = × 1 4 2 0 pe l R 13. Con sider el e men tary por tion of the rod of length dl at a dis tance l from the cen tre O of the rod. Charge on this portion dq dl Q L dl = = l \ dE dq a = × f 1 4 0 2 pe ( ) sec = × f 1 4 0 2 2 pe Q dl La sec Now, l a = f tan Þ dl a d = f f sec 2 \ dE Qd La = × f 1 4 0 pe Net Electric field at P. E dE = f ò cos [ dEsin f components will cancel each other as rod in symmetrical about P.] = × f f - ò 1 4 0 pe q q Q La d cos = × 1 4 2 0 pe q Q La sin But sinq= + æ è ç ö ø ÷ = + L a L L a L 2 2 4 2 2 2 2 \ E Q a a L = × + 1 4 2 4 0 2 2 pe 38 dE dE cos f dE cos f dE dE sin f dE sin f O f R df dl dl dE dE cos f dE cos f dE dE sin f dE sin f df Q P dl l dl f O a O a F e F e mg T sin a mg r T cos a T T 14. (a) As shown in fig ure, di rec tion of elec tric field at P will be along + ve y-axis. (b) Positive x-axis. (c) Positive y-axis. 15. Let E q R 1 0 2 1 4 = × pe Resultant fields of two opposite charges can be shown as given in figure. Clearly resultant field is along angle bisector of field towards 9 and 10. Hence time shown by clock in the direction of electric field is 9 : 30. 16. (a) a F m eE m = = - = - ´ ´ ´ ´ - - 1.6 9.1 10 1 10 10 19 3 31 = - ´ 1.76 10 14 ms -2 u = ´ 5.00 10 8 cm/s = ´ 5 10 6 ms -1 v = 0 v u as 2 2 2 - = s = ´ ´ ´ = ´ = - ( ) 5 10 2 10 1 10 6 2 14 2 1.7 .4 1.4 cm (b) v u at = + t = ´ ´ = ´ = - 5 10 10 10 28 6 14 8 1.76 2.8 ns. (c) Dk = work done by electric field. = × = - F x eEx = - ´ ´ ´ ´ ´ - - 1.6 10 1 10 8 10 19 3 3 = - ´ - 1.28 J 10 18 Loss of KE = ´ - 1.28 J 10 18 17. Here, u u x = ° = cos45 25 2 ms -1 u u y = ° = sin45 25 2 ms -1 a qE x = = ´ ´ ´ - 2 10 2 10 6 7 =40 ms -1 a y = -10 ms -1 y u a t yt y = + 1 2 y t t = - 25 2 5 2 39 E 1 E E 2 P x y –Q Q y +Q +Q E 2 E 1 E x P –Q +Q E 1 E 2 E P 12 E 12 2 3 4 5 6 7 8 9 10 11 E 1 E 2 E 3 E 4 E 5 E 6 E 7 E 8 E 9 E 10 E 11 1 6E 1 6E 1 6E 1 6E 1 E 6E 1 6E 1 q u E at the end of motion, t T = and y = 0 \ T = 5 2 s Also at the end of motion, x R = \ x u t a t x x = + 1 2 2 R= ´ + ´ æ è ç ö ø ÷ 25 2 5 2 20 5 2 2 =312.5 m 18. (a) R qE = m q 2 2 sin sin2 2 q= qER mu = ´ ´ ´ ´ ´ ´ ´ - - - 1.6 1.27 1.67 9.55 10 720 10 10 10 19 3 27 3 2 ( ) =0.96 2 88 q= ° or 92° q= ° 44 or 46° T mh E = 2 2 sinq = ´ ´ ´ ´ ´ ´ ´ - - 2 10 1 2 10 10 720 3 31 19 9.55 1.67 1.6 = ´ - 1.95 10 11 s 19. (a) a E j ® ® - - = - = - ´ ´ ´ e m 1.6 9.1 10 120 10 19 31 ^ = - ´ 2.1 10 13 i ^ m/s (b) t x u x = = ´ ´ = ´ - - D 2 10 10 4 3 10 2 5 7 1.5 s v u a t y y y = + = ´ ´ ´ ´ ´ - 3.0 2.1 10 10 4 3 10 6 13 7 = ´ 0.2 10 6 m/s v i j ® = ´ + ´ ( ) ( ) ^ ^ 1.5 0.2 10 10 5 6 20. Ab so lute po ten tial can be zero at two points on the x-axis. One in be tween the charges and other on the left of charge a 1 (smaller in mag ni tu de). Case I. In between two charges : let potential is zero at a distance x from q 1 towards q 2 . V q x q x = × + × - = 1 4 1 4 100 0 0 1 0 2 pe pe = × ´ - × ´ - = - - 1 4 2 10 1 4 3 10 100 0 0 6 0 6 pe pe x x Þ 200 2 3 - = x x x =20 cm Case II. Consider the potential is zero at a distance x from charge q, on its left. \ V q x q x = × + × + = 1 4 1 4 100 0 0 1 0 2 pe pe = × ´ + × - ´ + = - - 1 4 2 10 1 4 3 10 100 0 0 6 0 6 pe pe x x 200 2 3 + = x x x =200 cm 21. Let us first find the po ten tial at a point on the per pen dic u lar bi sec tor of a line charge. Consider a line of carrying a line charge density l having length L. Consider an elementary portion of length dl on the rod. Charge on this portion dq dl = l \ dV dl r = × f 1 4 0 pe l sec Now, l r = f tan dl r d = f f sec 2 40 O 100cm q 1 q 2 X q 1 q 2 x 100–x q 1 q 2 x 100 cm f q L r l dl \ dV d = f f l pe sec 4 0 \ V dV d = = × f f ò ò - l pe q q 4 0 2 sec = + - l pe q q q q 4 0 [ln| tan |] sec = + - ½ ½ ½ ½ ½ ½ é ë ê ê ù û ú ú l pe q q q q 4 0 ln tan tan sec sec = + 2 4 0 l pe q q ln| tan | sec In the given condition q = ° 60 Potential due to one side V V V 1 2 3 0 2 4 60 60 = = = × ° + ° l pe ln| tan | sec = × + 2 4 2 3 0 l pe ln| | Total potential at O V V = = × + 3 6 4 2 3 1 0 l pe ln| | = × + Q a 2 2 3 0 pe ln| | 22. (a) V V 2 1 2 250 20 10 - = - × = - ´ ´ ® ® - E d = - 50 V W V q V V = = - D ( ) 2 1 = ´ ´ - - 12 10 50 6 = -0.6 mJ (b) V V 2 1 50 - = - V 23. By work en ergy the o rem W K = D q V V mv mv ( ) 1 2 2 2 1 2 1 2 1 2 - = - - ´ - - 5 10 20 800 6 ( ) = ´ ´ - - 1 2 2 10 5 4 2 2 2 ( ( ) ) V v 2 2 55 = v 2 55 = = 7.42 ms -1 When a particle is released in electric field it moves in such a way that, it decreases its PE and increases KE Hence, particle at B is faster than that at A. 24. Cen tr e of cir cle is equi dis tant from ev ery point on its pe riph ery, Hence, V q R 0 0 1 4 = × pe , where q Q Q SQ = + = - 1 2 \ V Q R 0 0 1 4 5 = - × pe Similarly, V q R Z p = × + 1 4 0 2 2 pe = - × + 1 4 0 2 2 pe SQ R Z 25. Ini tial PE U q q r i = × 1 4 0 1 2 1 p e U q q r f = × 1 4 0 1 2 2 p e Work done by electric force W U U U f i = - = - - D ( ) = - × - æ è ç ç ö ø ÷ ÷ 1 4 1 1 0 1 2 2 1 pe q q r r Þ W = - ´ ´ ´ ´ - ´ - - 9 10 10 10 9 6 6 2.4 4.3 ( ) 1 1 0.25 2 0.15 - æ è ç ö ø ÷ W = -0.356 mJ 26. (a) U q q r q q r q q r = × + + é ë ê ê ù û ú ú 1 4 0 1 2 12 2 3 23 3 1 31 pe = ´ ´ ´ - ´ é ë ê ê - - 9 10 4 10 3 10 9 9 9 ( ) 0.2 + - ´ ´ ´ - - ( ) ( ) 3 10 2 10 9 9 0.1 + ´ ´ ´ ù û ú ú - - 4 10 2 10 9 9 0.1 U = ´ - - + = - - 9 10 6 6 8 360 8 [ ] nJ (b) Let the distance of q 3 from q 1 is x cm. Then U q q q q x q q x = × + - + é ë ê ê ù û ú ú = 1 4 0 0 1 2 2 3 3 1 pe 0.2 0.2 Þ 9 10 4 10 3 10 20 10 9 9 9 0 2 ´ ´ ´ - ´ ´ é ë ê ê - - - ( ) + - ´ ´ ´ - ´ - - - ( ) ( ) 3 10 2 10 20 10 9 9 2 x 41 O Page 5 8. As ball are in equi lib rium F T e = sin a mg T = cos a F mg e = tan a q r 2 0 2 4 = pe a tan Here, r l =2 sina q l 2 0 2 2 16 = pe a a sin tan q = ´ - 3.3 10 8 C. 9. Same as Q.7. In tro duc tory Ex er cise 21.3. 10. See Q.7. In tro duc tory Ex er cise 21.3. 11. E q r = × ® 1 4 1 0 3 pe r = ´ ´ - ´ + - - 9 10 10 9 9 2 232 ( ) (( ) ( )) ( ) / ^ ^ 8.0 1.2 1.6 1.2 1.6 i j =- - 18 2( ) ^ ^ 1.2 1.6 i j N/C. 12. Con sider an el e men tary por tion on the ring of length dl sub t end ing an gle df at cen tre ‘O’ of the ring. Charge on this portion, dq dl Rd = = f l l \ dE dq R d R = × = f 1 4 1 4 0 2 0 pe pe l Here, dE sin f components of field will cancel each other. Hence, Net field at O E dE R d = f = × f f ò ò - cos cos / / 1 4 0 2 2 pe l p p = × 1 4 2 0 pe l R 13. Con sider el e men tary por tion of the rod of length dl at a dis tance l from the cen tre O of the rod. Charge on this portion dq dl Q L dl = = l \ dE dq a = × f 1 4 0 2 pe ( ) sec = × f 1 4 0 2 2 pe Q dl La sec Now, l a = f tan Þ dl a d = f f sec 2 \ dE Qd La = × f 1 4 0 pe Net Electric field at P. E dE = f ò cos [ dEsin f components will cancel each other as rod in symmetrical about P.] = × f f - ò 1 4 0 pe q q Q La d cos = × 1 4 2 0 pe q Q La sin But sinq= + æ è ç ö ø ÷ = + L a L L a L 2 2 4 2 2 2 2 \ E Q a a L = × + 1 4 2 4 0 2 2 pe 38 dE dE cos f dE cos f dE dE sin f dE sin f O f R df dl dl dE dE cos f dE cos f dE dE sin f dE sin f df Q P dl l dl f O a O a F e F e mg T sin a mg r T cos a T T 14. (a) As shown in fig ure, di rec tion of elec tric field at P will be along + ve y-axis. (b) Positive x-axis. (c) Positive y-axis. 15. Let E q R 1 0 2 1 4 = × pe Resultant fields of two opposite charges can be shown as given in figure. Clearly resultant field is along angle bisector of field towards 9 and 10. Hence time shown by clock in the direction of electric field is 9 : 30. 16. (a) a F m eE m = = - = - ´ ´ ´ ´ - - 1.6 9.1 10 1 10 10 19 3 31 = - ´ 1.76 10 14 ms -2 u = ´ 5.00 10 8 cm/s = ´ 5 10 6 ms -1 v = 0 v u as 2 2 2 - = s = ´ ´ ´ = ´ = - ( ) 5 10 2 10 1 10 6 2 14 2 1.7 .4 1.4 cm (b) v u at = + t = ´ ´ = ´ = - 5 10 10 10 28 6 14 8 1.76 2.8 ns. (c) Dk = work done by electric field. = × = - F x eEx = - ´ ´ ´ ´ ´ - - 1.6 10 1 10 8 10 19 3 3 = - ´ - 1.28 J 10 18 Loss of KE = ´ - 1.28 J 10 18 17. Here, u u x = ° = cos45 25 2 ms -1 u u y = ° = sin45 25 2 ms -1 a qE x = = ´ ´ ´ - 2 10 2 10 6 7 =40 ms -1 a y = -10 ms -1 y u a t yt y = + 1 2 y t t = - 25 2 5 2 39 E 1 E E 2 P x y –Q Q y +Q +Q E 2 E 1 E x P –Q +Q E 1 E 2 E P 12 E 12 2 3 4 5 6 7 8 9 10 11 E 1 E 2 E 3 E 4 E 5 E 6 E 7 E 8 E 9 E 10 E 11 1 6E 1 6E 1 6E 1 6E 1 E 6E 1 6E 1 q u E at the end of motion, t T = and y = 0 \ T = 5 2 s Also at the end of motion, x R = \ x u t a t x x = + 1 2 2 R= ´ + ´ æ è ç ö ø ÷ 25 2 5 2 20 5 2 2 =312.5 m 18. (a) R qE = m q 2 2 sin sin2 2 q= qER mu = ´ ´ ´ ´ ´ ´ ´ - - - 1.6 1.27 1.67 9.55 10 720 10 10 10 19 3 27 3 2 ( ) =0.96 2 88 q= ° or 92° q= ° 44 or 46° T mh E = 2 2 sinq = ´ ´ ´ ´ ´ ´ ´ - - 2 10 1 2 10 10 720 3 31 19 9.55 1.67 1.6 = ´ - 1.95 10 11 s 19. (a) a E j ® ® - - = - = - ´ ´ ´ e m 1.6 9.1 10 120 10 19 31 ^ = - ´ 2.1 10 13 i ^ m/s (b) t x u x = = ´ ´ = ´ - - D 2 10 10 4 3 10 2 5 7 1.5 s v u a t y y y = + = ´ ´ ´ ´ ´ - 3.0 2.1 10 10 4 3 10 6 13 7 = ´ 0.2 10 6 m/s v i j ® = ´ + ´ ( ) ( ) ^ ^ 1.5 0.2 10 10 5 6 20. Ab so lute po ten tial can be zero at two points on the x-axis. One in be tween the charges and other on the left of charge a 1 (smaller in mag ni tu de). Case I. In between two charges : let potential is zero at a distance x from q 1 towards q 2 . V q x q x = × + × - = 1 4 1 4 100 0 0 1 0 2 pe pe = × ´ - × ´ - = - - 1 4 2 10 1 4 3 10 100 0 0 6 0 6 pe pe x x Þ 200 2 3 - = x x x =20 cm Case II. Consider the potential is zero at a distance x from charge q, on its left. \ V q x q x = × + × + = 1 4 1 4 100 0 0 1 0 2 pe pe = × ´ + × - ´ + = - - 1 4 2 10 1 4 3 10 100 0 0 6 0 6 pe pe x x 200 2 3 + = x x x =200 cm 21. Let us first find the po ten tial at a point on the per pen dic u lar bi sec tor of a line charge. Consider a line of carrying a line charge density l having length L. Consider an elementary portion of length dl on the rod. Charge on this portion dq dl = l \ dV dl r = × f 1 4 0 pe l sec Now, l r = f tan dl r d = f f sec 2 40 O 100cm q 1 q 2 X q 1 q 2 x 100–x q 1 q 2 x 100 cm f q L r l dl \ dV d = f f l pe sec 4 0 \ V dV d = = × f f ò ò - l pe q q 4 0 2 sec = + - l pe q q q q 4 0 [ln| tan |] sec = + - ½ ½ ½ ½ ½ ½ é ë ê ê ù û ú ú l pe q q q q 4 0 ln tan tan sec sec = + 2 4 0 l pe q q ln| tan | sec In the given condition q = ° 60 Potential due to one side V V V 1 2 3 0 2 4 60 60 = = = × ° + ° l pe ln| tan | sec = × + 2 4 2 3 0 l pe ln| | Total potential at O V V = = × + 3 6 4 2 3 1 0 l pe ln| | = × + Q a 2 2 3 0 pe ln| | 22. (a) V V 2 1 2 250 20 10 - = - × = - ´ ´ ® ® - E d = - 50 V W V q V V = = - D ( ) 2 1 = ´ ´ - - 12 10 50 6 = -0.6 mJ (b) V V 2 1 50 - = - V 23. By work en ergy the o rem W K = D q V V mv mv ( ) 1 2 2 2 1 2 1 2 1 2 - = - - ´ - - 5 10 20 800 6 ( ) = ´ ´ - - 1 2 2 10 5 4 2 2 2 ( ( ) ) V v 2 2 55 = v 2 55 = = 7.42 ms -1 When a particle is released in electric field it moves in such a way that, it decreases its PE and increases KE Hence, particle at B is faster than that at A. 24. Cen tr e of cir cle is equi dis tant from ev ery point on its pe riph ery, Hence, V q R 0 0 1 4 = × pe , where q Q Q SQ = + = - 1 2 \ V Q R 0 0 1 4 5 = - × pe Similarly, V q R Z p = × + 1 4 0 2 2 pe = - × + 1 4 0 2 2 pe SQ R Z 25. Ini tial PE U q q r i = × 1 4 0 1 2 1 p e U q q r f = × 1 4 0 1 2 2 p e Work done by electric force W U U U f i = - = - - D ( ) = - × - æ è ç ç ö ø ÷ ÷ 1 4 1 1 0 1 2 2 1 pe q q r r Þ W = - ´ ´ ´ ´ - ´ - - 9 10 10 10 9 6 6 2.4 4.3 ( ) 1 1 0.25 2 0.15 - æ è ç ö ø ÷ W = -0.356 mJ 26. (a) U q q r q q r q q r = × + + é ë ê ê ù û ú ú 1 4 0 1 2 12 2 3 23 3 1 31 pe = ´ ´ ´ - ´ é ë ê ê - - 9 10 4 10 3 10 9 9 9 ( ) 0.2 + - ´ ´ ´ - - ( ) ( ) 3 10 2 10 9 9 0.1 + ´ ´ ´ ù û ú ú - - 4 10 2 10 9 9 0.1 U = ´ - - + = - - 9 10 6 6 8 360 8 [ ] nJ (b) Let the distance of q 3 from q 1 is x cm. Then U q q q q x q q x = × + - + é ë ê ê ù û ú ú = 1 4 0 0 1 2 2 3 3 1 pe 0.2 0.2 Þ 9 10 4 10 3 10 20 10 9 9 9 0 2 ´ ´ ´ - ´ ´ é ë ê ê - - - ( ) + - ´ ´ ´ - ´ - - - ( ) ( ) 3 10 2 10 20 10 9 9 2 x 41 O + ´ ´ ´ ´ ù û ú ú = - - - 2 10 4 10 10 0 9 9 2 x Þ - - - + = 6 10 6 20 8 0 x x Þ x = 6.43 cm 27. Let Q be the third charge U q d qQ d qQ d = × + + é ë ê ê ù û ú ú = 1 4 0 0 2 pe Q q = - 2 28. V = - × ® ® E r (a) r k ® =5 ^ V = - - - = ( ) ( ) ^ ^ ^ 5 3 5 0 i j k (b) r i k ® = + 4 3 ^ ^ V = - - - + ( ) ( ) ^ ^ ^ ^ 5 3 4 3 i j i j = -20 kV 29. E ® = 400 j ^ V/m (a) r j ® = 20 ^ cm = ( ) ^ 0.2 j m V = - × = - ® ® E r 80 V (b) r j ® = - ( ) ^ 0.3 m V = - × = ® ® E r 120 V (c) r k ® = ( ) ^ 0.15 V = 0 30. E i ® = 20 ^ N/C (a) r i j ® = + ( ) ^ ^ 4 2 m V = - × = - ® ® E r 80 V (b) r i j ® = + ( ) ^ ^ 2 3 m V = - × = - ® ® E r 40 V 31. (a) [ [ ] [ ] A V xy yz zx ] = + + = - - [ ] [ ] ML T I L 2 3 1 2 = - - [ ] ML T I 0 3 1 (b) E V v x v y v z = - Ñ = - ¶ ¶ + ¶ ¶ + ¶ ¶ æ è ç ç ö ø ÷ ÷ ® i j k ^ ^ ^ = - + + + + + A y z z x x y [( ) ( ) ( ) ] ^ ^ ^ i j k (c) at (1m, 1m, 1m) E = - + + 10 2 2 2 ( ) ^ ^ ^ i j k = - + + 20( ) ^ ^ ^ i j k 32. V V B - = - × ® ® 0 E r Þ V - =- + 0 40 60 ( ) Þ V =-100 33. (a) E v x Ay Bx x = - ¶ ¶ = - - ( ) 2 E V y Ax C y - ¶ ¶ = - + ( ) E V Z z = - ¶ ¶ =0 (b) For E = 0 E x = 0 and E y = 0 Hence, E y = 0 Ax C + = 0 x C A = - E x = 0 Ay B C A - - æ è ç ö ø ÷ = 2 0 y BC A = - 2 2 Hence, E is zero at - - æ è ç ö ø ÷ C A BC A , 2 2 . 34. f = q e 0 q = = ´ ´ - e 0 12 10 360 f 8.8 = ´ - 3.18 10 9 C =3.186 nC 36. (a) f= = - ´ ´ - - q e 0 6 12 10 10 3.60 8.85 = ´ 4.07 10 5 V-m. (b) f = q e 0 Þ q = e 0 f = ´ ´ = ´ - - 8.85 6.903 10 780 10 12 9 q =6.903 nC (c) No. Net flux through a closed surface does not depend on position of charge. 36. E i j ® = + æ è ç ö ø ÷ 3 5 4 5 0 0 E E ^ ^ S j ® = 0.2 ^ m 2 = 1 5 j ^ m 2 \ f = × = ® ® E S 4 25 Nm 2 /C 42Read More
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