Downloe fo HPTER? ELETRI FIELD ND POTENTIL EXERISES. oulob Newton l M L T 4 k F.. istnce between k so, foce k ( F ( The weight of boy 4 N 4 N wt of boy.5 So, foce between chges 4 So, foce between chges 5.6 weight of boy.., Let the istnce be F 5.8 4 K F 4 o 4 4. 4. chges?? ech, wt, of 5 kg peson 5 g 5.8 4 N k k F 4 N 4 4 6,5 N (5.6 5.6 8.6 6 54.4. 5 coulob. 4 coulob 5. hge on ech poton.6 coulob Distnce between chges 5 ete 8 k.6.6 Foce.56.4 Newton 6.. 6. 6 c. Let the chge be t istnce fo F K k F (. 6. (.? c Now since the net foce is zeo on the chge. f f k k (. (. (...586 5.86 c 5. c Fo lge chge. Downloe fo
Downloe fo Electic Fiel n Potentil 7. 6 c? 6 c c Let the thi chge be so, F-? F- k K 6 ( ( + + ( - 4.4 c.44 So, istnce 4.4 + 4.4 c fo lge chge 8. Miniu chge of boy is the chge of n electon Wo,.6? c c c So, F k.6.6 6.4?8+++.4 5. 4. No. of electons of g wte 55.5 Nos Totl chge 55.5 8 No. of electons in 8 g of H O 6. 6. 4 6. No. of electons in g of H O.4 6.4 5 8 Totl chge.4 5.6 5.4 6 c. Molecul weight of H O 6 6 No. of electons pesent in one olecule of H O 8 g of H O hs 6. olecule 8 g of H O hs 6. electons 4 4 6. g of H O hs electons 8 So nube of potons hge of potons 6. 8.6 6 6. 8 potons (since to is electiclly neutl 6 coulob.6 6. 8.6 6. hge of electons coulob 8 7.6 6..6 6. 8 8 Hence Electicl foce ( 7 8 6. 5.6 6..56 5 Newton 8. Let two potons be t istnce be.8 fei F.6 (4.8 8. N 7 7 coulob. F. N c (s they ubbe with ech othe. So the chge on ech sphee e eul So, F k. k (. 4 4.6 c ies by electon c cie by.6 6 c 6 c 7 + ++ + +. 7 c cies by.6 7..8.8. Downloe fo
Downloe fo Electic Fiel n Potentil. F k.6.6 (.75.4 7.56.4 4. Given: ss of poton.67 7 kg M p k hge of poton.6 c p G 6.67 Let the seption be?? Fe k( p Now, Fe : Fg, fg K( p G(M p G(M 5. Epession of electicl foce F e p 6.67 k (.6 ` (.67 Since e k is pue nube. So, iensionl foule of F 7 O, [MLT ][L ] iensionl foule of [ML T ] Unit of unit of foce unit of Newton Newton? Since?k is nube hence iensionl foule of k [L ] Unit of k i entionl foule of 6. Thee chges e hel t thee cones of euiltel tngle. Let the chges be, n. It is of length 5 c o.5 Foce eete by on F foce eete by on F So, foce eete on esultnt F F k 6 F 4 4.4 5 5 5 Now, foce on F cos since it is euiltel. Foce on.44 7. 4 6 v 5 c 5? so foce on c F F F 4.4 N. so Foce long oponent F F cos 45 k( (5 6 4 4 4 D D k( (5 6 k 5 5.56 8,4 8 i ensionl foule of i ensionl foule of 4 4.44 (.5.4 Foce long % coponent.4 D.5 F F 6.5 F.5 F D F So, Resultnt R 8. R.5.5 F K F Fy.4 7.56.6.6.5.5 8 8. N. Fe fo pevious poble No. 8 8. 8 N Ve? Now, M e. kg.5 Mev Now, Fe v v.8 6 /s Fe 8..5. e 8..4775 4.775 /s Downloe fo
Downloe fo Electic Fiel n Potentil. Electic foce feele by c ue to 8 c. F F k ( k 8 ( 8 8 k -6 N. electic foce feele by c ue to 8 8 c. k 8 8 8k 4 6 k 6 N. 8 k 7 Siilly F k 6 N ( So, F F + F + F +??+ F k 6 ( + + +??+ N k 6 55k 6 55 6 N 4.5 N k. Foce eete 6.6 6 is the foce eete on the sting. 7 c g l 5 c 5 5 ( Now Electic foce T 4 T os 4 F K N 4.4 N.44 N 4 5 F (b The coponents of Resultnt foce long it is zeo, becuse blnces T cos n so lso. F T sin (c Tension on the sting T sin F T cos F.44 Tn. 8.46 ut T cos N T cos sec F T, sin Sin.456 ; os.878;.. 8 c n? T? Sin Foce between the chges 8 8 K F 4 N ( F 4 sin F 8 8 g gsin (/ os Sin 4 So, T cos O T 8. 8 M 4 4. T Sin T os c T c 5 c c c T Sin F T.4 Downloe fo
Downloe fo Electic Fiel n Potentil 4. T os ( T Sin Fe ( Fe Solving, (/( we get, tn 56 (.4..8 (.4..8 56 7 5. Electic foce 6 4. 8 c k ( sin sin k 4 6.7 7 6 c.5 4 k sin So, T os s (Fo euilibiu T sin Ef Ef O tn Ef cot cot o unit. 6E Sin g k 4 sin cot cot sin 6E 6. Mss of the bob g. kg So Tension in the sting..8.8 N. Fo the Tension to be, the chge below shoul epel the fist bob. k F T? + F T? f T 4.8.8.54 N 5 (. 7. Let the chge on So, net foce on c is eul to zeo So F F, ut F F k k ( (?? ( ( ( ( Fo the chge on est, F + F k k( k (.44 [(.44 ]?(.44 (?(.4?(6? 4 8. K N/ l c. 8 c Fin l? Foce between the F So, F? k o F K k 6 5 8.5 8 6 7 c.6 6 E F 6 5 N g l v l sin 4 c l 56 g 4 c FD fo ss T cos ( T E F T Sin c 4 K Downloe fo
Downloe fo Electic Fiel n Potentil. 6 M b 8 g. Since is t euilibiu, So, Fe R K R g 6...8.8..8.8. 8.7 8 Rnge 8.7 8 6. 6 c Let the istnce be unit F epulsion Fo euilibiu k k 4 8 4.8 sin.8 7.8 7.4 ete.74 ete fo the botto.. Foce on the chge pticle?? t?c? is only the coponent of foces So, F on c F Sin + F Sin ut F F F Sin K ( / F Fo iu foce 6k (4 / / 4 / ( k / / 4 / (4 / 4 8 K [4 ] / K(4 (4 (4 (4 + 6 4 + 4 + 8 4 + 8 + 8 8. ( Let chge on & Septe by istnce chge on c isplce to So, foce on F F ut F O os F O os So, foce on?? in ue to veticl coponent. O F F O Sin + F O Sin K Sin ( / 4 k ( 4 [( / ] / FO FO F K ( / Sin 6k (4 / k Electic foce F / [( / ] R F F O / c R / O F F O F e.6 Downloe fo
Downloe fo Electic Fiel n Potentil k (b When << F [( / / ] F ( k / 4 / So tie peio T g. F K ( << F k F [( / 4 F Net foce K ( K ( ( ( ( 4 K K ( ( ( <<< l / neglecting w..t. l K4 K4 net F 4 We get cceletion isplceent Tie peio cceletion 4K 4K 4K 4 4 4 4 8 4 4. F e.5 N, 6, F e E E F e.5 6.5 N/ 5. 6,? 4 6, c. (E electic fiel ue to, E electic fiel ue to ( 6. EF.44 K 5 N/ 4 (. 44 4 8.85 c.44 8.88? 4 6 6.44 +.44 7., g kg,.5 6 ut E (.5 6 E 6 E.5 4 6.5 66.6 N/.5 5 6.6 5 / l l+ X l l F os 6 E.7 Downloe fo
Downloe fo Electic Fiel n Potentil 8.. 8, l c E? V? Since it fos n euipotentil sufce. So the electic fiel t the cente is Zeo. ( ( 4 (4. 7.5 8 V. V. We know : Electic fiel?e? t?p? ue to the chge ing K K / (R R K Foce epeience?f? E R Now, plitue R 4 R So, T K / R K T 6 R / 4. hge pe unit length L fo length l l 4 4R Electic fiel t the cente ue to chge k The hoizontl oponents of the Electic fiel blnces ech othe. Only the veticl coponents ein. Net Electic fiel long veticl K cos kos E E cos l [but l ] k k os os / o E but L R L / k os k Sin k l K L. 8. 8. 8 R O X l l P, l L So E k L (L / k L 4 L L 4. G 5 5 6 K We hve, E fo chge cyline. 6 5 5 E. 5 5 5 + + + + + 5.8 Downloe fo
Downloe fo Electic Fiel n Potentil 4. Electic fiel t ny point on the is t istnce fo the cente of the ing is E 4 / (R (R Diffeentiting with espect to / K / E K(R K( / (R ( / Since t istnce, Electic fiel is iu. R E K (R + /? K (R + / K (R + / K (R + / R + R R R 4. Since it is egul hegon. So, it fos n euipotentil sufce. Hence the chge t ech point is eul. Hence the net entie fiel t the cente is Zeo. 44. hge/unit length hge of Initilly the electic fiel ws?? t the cente. Since the eleent? is eove so, net electic fiel ust K Whee chge of eleent l K E 4 8 45. We know, Electic fiel t point ue to given chge K E Whee chge, Distnce between the point n the chge So, E [?? hee] 4 46. E kv/ v/, 8 5 kg, c 5 tn E [ T Sin, T os e] 8 tn 6 8 + tn 5 4 4 [os T Sin T 8 6 5 4 8 T 47. Given 5 4, Sin ] 5 5 4 5 8. 4 u Velocity of pojection, E Electic fiel intensity hge; ss of pticle We know, Foce epeience by pticle with chge?? in n electic fiel E E E cceletion pouce T E E E. Downloe fo
Downloe fo s the pticle is pojecte ginst the electic fiel, hence eceletion E Electic Fiel n Potentil So, let the istnce covee be?s' Then, v u + s [whee cceletion, v finl velocity] E Hee u u S S units E 48. g kg, u,.5 4 ; E. 4 N/c ; S 4 c 4 F E.5 4. 4 N F So, E.8.8 N b S t o t g 4.6 sec v u + s + 4 4 v 4 4. 4 /sec wok one by the electic foce w Ft 4. J 4. g, 4. 5, F g, F e E E 4 N/ So, the pticle oves ue to the et esultnt R R g e F F e 5 (..8 (4..64 6.4. 8.85 N F g tn So, 45 F E Hence pth is stight long esultnt foce t n ngle 45 with hoizontl Disp. Veticl (/.8.6 Disp. Hoizontl S (/ t Net Dispt. 5. 4 g, 4 6 4 E.8 t.6. (.6 (.6 768. 7.7 45 Tie fo oscilltions 45 sec. Tie fo oscilltion sec 45 When no electic fiel is pplie, T g 45 4 l (45 ( 4.86 When electic fiel is not pplie, E.86 T [.5].58 g. 5 Tie fo oscilltion.58 Tie fo oscilltion.58 5.6 sec 5 sec. 5. F E, F K Whee plitue E E? K o K K 45 E E R E. Downloe fo
Downloe fo Electic Fiel n Potentil 5. The block oes not unego. SHM since hee the cceletion is not popotionl to isplceent n not lwys opposite to isplceent. When the block is going tows the wll the cceletion is long isplceent n when going wy fo it the isplceent is opposite to cceletion. Tie tken to go tows the wll is the tie tken to goes wy fo it till velocity is ut + (/ t E t t E t E Totl tie tken fo to ech the wll n co bck (Tie peio t E 8 E 5. E n/c, S 5 c. V E o, V E.5 5 c 54. Now, V? V Potentil iff? hge. Wok one J Now, Wok one Pot. Diff hge Pot. Diff Volt. 55. When the chge is plce t, K K4 E 7 7 ( (.. 4 4. 7 4 J When chge is plce t, 4 K K4 4 E. 6 4 J Wok one E? E (7? 6 4 6 4 J.6 J 56. ( (, (4, V? V E 6 8 V (b (4,, (6, 5 V? V E (6 4 4 V (c (, (6, 5 V? V E (6 6 V. 57. ( The Electic fiel is long -iection Thus potentil iffeence between (, n (4, is, V E? (4? 8 V Potentil enegy (U? U between the points V? 8 (? 4 6 4.6 J. (b (4, (6, 5 V? E?? 4 V Potentil enegy (U? U between the points V? 4 (? 4 8 4.8 J (c (, (6, 5 V? E? 6? V Potentil enegy (U? U between the points n V? (? 4 4 4.4 J. 7 c z 7 y c E N/ Downloe fo
Downloe fo 58. E ĵ î N/V t (, ( i + j So, V E?(i + J ( î + j?( +? V 5. E i i v v E V? V? 5 V 5 Volts 6. V(, y, z (y + yz + z ( (b E Volt ML T TL [MT ]. Electic Fiel n Potentil Vî Vĵ Vkˆ [(y yz z [(y yz z [(y yz z y z y z ( y zî ( zĵ (y kˆ (y zî ( zĵ (y kˆ (c SI unit, (,, E ( î? ( ĵ? ( kˆ? î? ĵ? kˆ o 4.64 5 N/ 6. 5 Ech e bought fo infinity to c pt So wok one negtive of wok one. (Potentil E 4 P.E F s P.E. K 6 J 6. ( The ngle between potentil E l v Y E v v v 4 v hnge in potentil V V s E V (s potentil sufce So, E l V E l os( +? v 4 E( cos? V V E V/ king n ngle with y-is os ( / v v (b s Electic fiel intensity is to Potentil sufce v 6 v k k k 6 So, E 6 v K v v k 6 k 6 So, E v. v. k 6. Rius So, icufeence hge ensity Totl chge K Electic potentil 4 / / ( ( V So, Electic fiel os / ( ( / / / ( ( / ( O Y P(, ( X Downloe fo
Downloe fo Electic Fiel n Potentil 64. E N/ ( V E l V (b u? E, / F E.6.75 4 /s. u?.75 4. u.4.75 4 u.64 6 /s. (c Now, U u os 6 V, s?.75 4 /s V u? s uos6 s 6.64 4.75.75.5 4.47.. c 65. E N/ in -iection ( Potentil t the oigin is O. V? E? E y y? E z z V?? V? (b (5???.5 (c If potentil t oigin is v, v?? V? +? ( Potentil t IS, V? V? V V + + V Potentil t oigin is. No, it is not pcticl to tke potentil t to be zeo. 66. ount of wok one is ssebling the chges is eul to the net potentil enegy 5 So, P.E. U + U + U c c K K K K [4 4 ] 6 (8 6 6 4 J 67. K.. eceses by J. Potentil v to v. So, chnge in K.E ount of wok one J (? v v. 68. g; F K 4 F.8 7 7.8 F.8 /s V? u s V u + s V.8.6.6 6 /s. 6. 4 5 ; s, 5 g.5 kg 5 (4 F K 4.4 N +4 5 F 4.4 cceletion?? 88 /s. 5 Now u, s 5 c.5, 88 /s, V? V u + s V 88.5 V 88 5.66 /s 4 /s fo ech pticle 4 O 4 c E 4 5 c u cos 6 E 6 c 5 O 4 c? 4 5. Downloe fo
Downloe fo Electic Fiel n Potentil 7. E.5 4 P.4 PE sin P E.4.5 4 8.5 6 7. ( Dipoleoent l (Whee gnitue of chge 6 - c 8 c (b We know, Electic fiel t n il point of the ipole 8 KP 6 7 N/ ( Seption between the chges 6 Hence length so, Tie peio (E / E 75. 64 gs of coppe hve ole 6.4 gs of coppe hve. ole ole No tos. ole (no. tos 6. tos 6 tos to contibutes electon 6 tos contibutes 6 electons. O c c? 6 (c We know, Electic fiel t point on the pepenicul bisecto bout M wy fo cente of ipole. 8 KP 8 N/ O 7. Let? &? e plce t & Whee on So length of 6 So the ipole oent ( P So, Resultnt ipole oent P [( + ( + os 6 ] / [ ] / P 7. ( P E (b E Sin E sin Electic fiel intensity P E os + E os E os P P Kp KP K E so E / / E ( ( When << K PK P ( / 4 + 74. onsie the o to be siple penulu. Fo siple penulu T / g ( length, cceletion E Now, foce epeience by the chges F E F E Now, cceletion.4 Downloe fo