Tiêu đề Notes_Exercise_11.Electric Field And Potential.pdf ✅

29.1 CHAPTER – 29 ELECTRIC FIELD AND POTENTIAL EXERCISES 1. 0 = 2 2 Newton m Coulomb = l1 M–1L–3T4  F = 2 1 2 r kq q 2. q1 = q2 = q = 1.0 C distance between = 2 km = 1 × 103 m so, force = 2 1 2 r kq q F = 3 2 9 2( 10 ) 9( 10 ) 1 1     = 2 6 9 2 10 9 10   = 2,25 × 103 N The weight of body = mg = 40 × 10 N = 400 N So, force between charges wt of body = 1 2 3 4 10 .2 25 10            = (5.6)–1 = 6.5 1 So, force between charges = 5.6 weight of body. 3. q = 1 C, Let the distance be  F = 50 × 9.8 = 490 F = 2 2 Kq   490 = 2 9 2 9 10 1    or 2 = 490 9 109  = 18.36 × 106   = 4.29 ×103 m 4. charges ‘q’ each, AB = 1 m wt, of 50 kg person = 50 × g = 50 × 9.8 = 490 N FC = 2 1 2 r kq q  2 2 r kq = 490 N  q2 = 9 2 9 10 490 r   = 9 9 10 490 1 1     q = 9 54 4. 10  = 23.323 × 10–5 coulomb ≈ 2.3 × 10–4 coulomb 5. Charge on each proton = a= 1.6 × 10–19 coulomb Distance between charges = 10 × 10–15 metre = r Force = 2 2 r kq = 30 9 38 10 9 10 6.1 6.1 10       = 9 × 2.56 × 10 = 230.4 Newton 6. q1 = 2.0 × 10–6 q2 = 1.0 × 10–6 r = 10 cm = 0.1 m Let the charge be at a distance x from q1 F1 = 2 Kq1q  F2 = 2 2 1.0( ) kqq   = 2 6 9 9.9 2 10 10 q       Now since the net force is zero on the charge q.  f1 = f2  2 kq1q  = 2 2 1.0( ) kqq    2(0.1 – ) 2 = 2  2 (0.1 – ) =    = 1 2 1.0 2  = 0.0586 m = 5.86 cm ≈ 5.9 cm From larger charge q2 x m (0.1–x) m q q1 10 cm Page 1 ELECTRIC FIELD AND POTENTIAL
Electric Field and Potential 29.2 7. q1 = 2 ×10–6 c q2 = – 1 × 10–6 c r = 10 cm = 10 × 10–2 m Let the third charge be a so, F-AC = – F-BC  2 1 1 r kQq = 2 2 2 r KQq  2 6 (10 ) 2 10     = 2 6 1 10     22 = (10 + ) 2  2  = 10 +   ( 2 - 1) = 10   = .1 414 1 10   = 24.14 cm  So, distance = 24.14 + 10 = 34.14 cm from larger charge 8. Minimum charge of a body is the charge of an electron Wo, q = 1.6 × 10–19 c  = 1 cm = 1 × 10–2 cm So, F = 2 1 2 r kq q = 2 2 9 19 19 10 10 9 10 6.1 6.1 10 10           = 23.04 × 10–38+9+2+2 = 23.04 × 10–25 = 2.3 × 10–24 9. No. of electrons of 100 g water = 18 10 100 = 55.5 Nos Total charge = 55.5 No. of electrons in 18 g of H2O = 6.023 × 1023 × 10 = 6.023 × 1024 No. of electrons in 100 g of H2O = 18 .6 023 10 100 24   = 0.334 × 1026 = 3.334 × 1025 Total charge = 3.34 × 1025 × 1.6 × 10–19 = 5.34 × 106 c 10. Molecular weight of H2O = 2 × 1 × 16 = 16 No. of electrons present in one molecule of H2O = 10 18 gm of H2O has 6.023 × 1023 molecule 18 gm of H2O has 6.023 × 1023 × 10 electrons 100 gm of H2O has 100 18 .6 023 1024   electrons So number of protons = 18 .6 023 1026  protons (since atom is electrically neutral) Charge of protons = 18 6.1 10 .6 023 10 19 26     coulomb = 18 6.1 .6 023 107   coulomb Charge of electrons = = 18 6.1 .6 023 107   coulomb Hence Electrical force = 2 2 7 7 9 (10 10 ) 18 6.1 .6 023 10 18 6.1 .6 023 10 9 10                        = 25 6.1 .6 023 10 18 8 .6 023     = 2.56 × 1025 Newton 11. Let two protons be at a distance be 13.8 femi F = 2 30 9 38 (14 )8. 10 9 10 6.1 10       = 1.2 N 12. F = 0.1 N r = 1 cm = 10–2 (As they rubbed with each other. So the charge on each sphere are equal) So, F = 2 1 2 r kq q  0.1 = 2 2 2 (10 ) kq   q2 = 9 4 9 10 1.0 10     q2 = 14 10 9 1    q = 7 10 3 1   1.6 × 10–19 c Carries by 1 electron 1 c carried by 19 6.1 10 1   0.33 × 10–7 c carries by 7 19 .0 33 10 6.1 10 1      = 0.208 × 1012 = 2.08 × 1011 + + + – – – – + + A 10 × 10–10 m C B a 2 × 10–6 c –1 × 10–6 c Page 2 ELECTRIC FIELD AND POTENTIAL
Electric Field and Potential 29.3 13. F = 2 1 2 r kq q = 10 2 9 19 19 .2( 75 10 ) 9 10 6.1 6.1 10 10          = 20 29 .7 56 10 23.04 10     = 3.04 × 10–9 14. Given: mass of proton = 1.67 × 10–27 kg = Mp k = 9 × 109 Charge of proton = 1.6 × 10–19 c = Cp G = 6.67 × 10–11 Let the separation be ‘r’ Fe = 2 2 p r (k C ) , fg= 2 2 p r G(M ) Now, Fe : Fg = 2 p 2 2 2 p G(M ) r r K(C )  = 11 27 2 9 19 2 .6 67 10 .1( 67 10 ) 9 10 6.1( 0`1 )          = 9 × 2.56 × 1038 ≈ 1,24 ×1038 15. Expression of electrical force F = 2 r kr C e   Since e–kr is a pure number. So, dimensional formulae of F = 2 dimensional formulae of r dim ensional formulae of C Or, [MLT–2][L2 ] = dimensional formulae of C = [ML3 T–2] Unit of C = unit of force × unit of r2 = Newton × m2 = Newton–m2 Since –kr is a number hence dimensional formulae of k = dim entional formulae of r 1 = [L–1] Unit of k = m–1 16. Three charges are held at three corners of a equilateral trangle. Let the charges be A, B and C. It is of length 5 cm or 0.05 m Force exerted by B on A = F1 force exerted by C on A = F2 So, force exerted on A = resultant F1 = F2  F = 2 2 r kq = 4 9 12 5 5 10 9 10 2 2 2 10          = 10 25 36  = 14.4 Now, force on A = 2 × F cos 30° since it is equilateral .  Force on A = 2 × 1.44 × 2 3 = 24.94 N. 17. q1 = q2 = q3 = q4 = 2 × 10–6 C v = 5 cm = 5 × 10–2 m so force on c = FCA  FCB  FCD so Force along × Component = FCD  FCA cos 45  0 = 2 2 1 5( 10 ) 2(k 10 ) 5( 10 ) 2(k 10 ) 2 2 6 2 2 2 6 2          =            4 4 2 50 2 10 1 25 10 1 kq =                2 2 1 1 24 10 9 10 4 10 4 9 12 = 1.44 (1.35) = 19.49 Force along % component = 19.49 So, Resultant R = 2 2 Fx  Fy = 19.49 2 = 27.56 18. R = 0.53 A° = 0.53 × 10–10 m F = 2 1 2 r Kq q = 10 10 9 38 .0 53 .0 53 10 10 9 10 6.1 6.1 10           = 82.02 × 10–9 N 19. Fe from previous problem No. 18 = 8.2 × 10–8 N Ve = ? Now, Me = 9.12 × 10–31 kg r = 0.53 × 10–10 m Now, Fe = r M v 2 e  v 2 = me Fe  r = 31 8 10 1.9 10 2.8 10 .0 53 10        = 0.4775 × 1013 = 4.775 × 1012 m2 /s2  v = 2.18 × 106 m/s 0.05 m 60°  F2 F1 B A C 0.05 m 0.05 m D C B FCD A FCA FCB Page 3 ELECTRIC FIELD AND POTENTIAL