Odpowiedzi do zadań z zestawu 1E

1. a) x=L/[1+
]=0.5m, między q₁ i q₂ , b) x=L/[1-
]=2.7m, po stronie q₂

2. Q=-q(1/2+
)/2=-0.67nC 3. q=4Lsin(α/2)
=39.6nC

4. ε=ρ_k/(ρ_k-ρ_c)=1.5 5. F_||=qQL/[4ε₀(L²+R²)^3/2]=4.91N

6. V=qn^2/3/4πε₀r=899.2V 7. tgα=4πε₀UVr/mgd=0.113366, α=6.47^o

8. T_e=2π
=0.448s

9. a) E_x=k[q₂a/(a²+b²)^3/2-|q₁|/a²], E_y=k[q₂b/(a²+b²)^3/2-|q₃|/b²],

E=(E_x²+E_y²)^1/2=2.02·10⁵N/C, tgφ=E_y/E_x, φ=-74,3^o,

b) W=Q{(2q₂+2q₃-q₁)/a+q₁/[(a/2)²+b²]^1/2-q₃/b-q₂/(a²+b²)^1/2 }/4πε₀=1.67J

10. a) E_x=k[(q₁-q₂)/(b/2)²+|q₃|a/(2[(a/2)²+b²]^3/2)], E_y=-k|q₃|b/[(a/2)²+b²]^3/2,

E=(E_x²+E_y²)^1/2=3.24·10⁵N/C, tgφ=E_y/E_x, φ=-18.5^o,

b) W=Q[2(q₁+q₂+q₃)/(a²+b²)^1/2-2(q₁+q₂)/a-q₃/[(a/2)²+b²]^1/2/4πε₀=1.28J

11. V=(V₁R₁+V₂R₂)/(R₁+R₂)=260V ,

Odpowiedzi do zadań z zestawu 2E

12. v=(qQ/2πε₀Lm)^1/2=5.05m/s

13. E₁=E₂, E₁=2kqQ/[s²+(L/2)²]^1/2, E₂=2kqQ/[x²+(L/2)²]^1/2+mv²/2, v_max dla x=0,

v_max={2qQ[1/(4s²+L²)^1/2-1/L]/πε₀m}^1/2=4.02m/s ; F=2kq|Q|sinφ/[x²+(L/2)²],

sinφ=x/[x²+(L/2)²]^1/2, F≅16kq|Q|x/L³ dla x<<L, ma=-4q|Q|x/πε₀L³, x''+4q|Q|x/πε₀mL³=0,

ω²=4q|Q|x/πε₀mL³, T=2πωπL(πε₀mL/q|Q|)^1/2=0.04s

14. F=|(dE_p/dx)|, x=d, E_p=Q²/2C, C=ε₀S/x, E_p=Q²x/2ε₀S, F=Q²/2ε₀S=CU²/2d=1N

albo F=QE, E=σ/2ε₀, σ=Q/S, F=Q²/2ε₀S i dalej jw.

15. U<(C₁+C₂)U₁/C₂=107.2V, i U<(C₁+C₂)U₂/C₁=97V

16. a) C_ef=(C₁+C₂)C₃/(C₁+C₂+C₃)=3.53nF, b) Q₁=C₁U₁=UC₁C₃/(C₁+C₂+C₃)=2.51·10^-7C,

c) C₃→C₃+C₄, Q₁'=UC₁(C₃+C₄)/(C₁+C₂+C₃+C₄)=3.37·10^-7C, ΔQ=Q₁'-Q₁=0.86·10^-7C,

d) C_ef'=(C₁+C₂)(C₃+C₄)/(C₁+C₂+C₃+C₄)=4.75nF

17. zał.: V_B=V_min, U_MN=U[C₁/(C₁+C₂)-C₃/(C₃+C₄)]= -8.4V

lub U_MN=U[C₄/(C₃+C₄)-C₂/(C₁+C₂)]= -8.4V, C_ef=C₁C₂/(C₁+C₂)+C₃C₄/(C₃+C₄)=36nF

18. zał. Q₃>0, Q₃=(E₁C₁-E₂C₂)C₃/(C₁+C₂+C₃)=39.6nC, (U₃=1.2V),

gdy C₃=0 i E₂
-E₂ wtedy U_MN==(E₁C₁+E₂C₂)/(C₁+C₂)=10.7V

19. a) W=E₂-E₁, E₁=Q²/2C=CU²/2, E₂=Q²/2C', C'=C/n, E₂=nq²/2C=nCU²/2,

W=(n-1)CU²/2=5.4mJ, b) E₁+W=E₂+W_q, W=E₂-E₁+W_q, E₂=C'U²/2, C' jw.,

W_q=ΔQU, ΔQ=|Q'-Q|=CU-C'U=C(1-1/n)U, W=(n-1)CU²/2n=1.35mJ

20. a) C'=C/ε, W=(ε-1)CU²/2=12mJ, b) W=(ε-1)CU²/2ε=2.4mJ

21. mv²/2=qU₁, vT=L, ma_x=F_e, F_e=qE=qU₂/d, x=aT²/2, a=qU₂/md, T=L/v, v²=2qU₁/m,

x=L²U₂/4dU₁=0.25cm, tgα=LU₂/2dU₁=0.025, α=1.4^o

Odpowiedzi do zadań z zestawu 3E

22. U₁=q/C₁=226.7V, U₁'=nC₁C₂²U/{(C₁+C₂)[C₁(C₂+C₃)+nC₂C₃]}=197.8V,

W₁=C₁U₁²/2=2.1μJ, W₁'=C'(U₁')²/2=0.803μJ, ΔW₁=W₁'-W₁=-1,3μJ,

E=U₁/d=5.67·10⁵V/m, E'=U₁'/nd=2.47·10⁵V/m, ΔE=E'-E=-3.19·10⁵V/m

23. C_x=εC₁C₂(U'-U)/[(C₁+C₂)(εU-U')]=29.9pF

24. Q_x'=C₁(U-U₃)/[1+C₂/ε(C₁U/U₃-C₁-C₂)]=6.55nC, W_x'=(Q_x') ²/2C_x'=0.178μJ,

E_x^'=C₁(U-U₃)/d[ε(C₁U/U₃-C₁-C₂)+C₂]=1.81·10⁵V/m

25. Q₃=C₁C₃U/(C₁+C₂+C₃)=8.56nC, W₃=Q₃²/2C₃=0.78μJ, E₃=U₃/d₁=Q₃/C₃d₁=1.82·10⁵V/m,

n=d₁/d₂, Q₃'=nC₁C₃(C₂+C₃)U/[(C₁+C₂+C₃)(C₂+nC₃)]=11.57nC, ΔQ₃=Q₃'-Q₃=3.01nC,

W₃'=(Q₃')²/2C₃'=0.713μJ, ΔW₃=W₃'-W₃=-0.067μJ,

E'=U₃'/d₂=Q₃'/C₃'d₂=Q₃/C₃d₁=2.46·10⁵V/m, ΔE=E'-E=6.41·10⁴V/m

26. I=2πRλf=1mA, 27. v=[|q|(Q-|q|/4)/(4πε_omR)]^1/2=1.9m/s, I=|q|v/πR=60nA

28. a) j=4I/πD²=3.18A/mm², b) n_e=ρN_A/μ=8.45^.10²⁸/m³, c) v_u=4IμπρeN_AD²=0.235mm/s

29. I=CUv(ε-1)/εL=90μA, 30. R_AB=[2R₁R₂+(R₁+R₂)R₃]/(R₁+R₂+2R₃)=39Ω

Odpowiedzi do zadań z zestawu 4EM

31. Q=CER₂/(R₁+R₂+r)=30.9nC

32. U_AB=E(R₃R₂-R₁R₄)/[(R₁+R₂)(R₃+R₄)+(R₁+R₂+R₃+R₄)r]=7.16V

33. U=(E₁r₂+E₂r₁)/(r₁+r₂)=5V

34. I₁=[(E₁-E₃)R₂+(E₁-E₂)R₃]/W=7.81mA, I₂=[(E₂-E₁)R₃+(E₂-E₃)R₁]/W=-1.09mA,
I₃=[(E₃-E₁)R₂+(E₃-E₂)R₁]/W=-6.72mA

35. W=(R₁+r)(R₂+r)+(R₁+R₂+2r)R₃, M=(P/R₃)^1/2, a) E₂<[WM-E₁(R₂+r)]/(R₁+r)=7.04V
albo E₂>[-WM-E₁(R₂+r)]/(R₁+r)=-20.4V (odwr. pol.), b) E₂>0 i I₂<0 →

E₂<(E₁R₃)/(R₁+R₃+r)=5.71V oraz I₁>0, → E₂<E₁(R₂+R₃+r)/R₃=17.5V czyli E₂<5.71V

36. R=mvsinφ/qB=6.0cm, T=2πm/qB=0.26μs, h=2πmvcosφ/qB=10.1cm

37. L=2mU/B=6.64·10^-21J·s, 38. n_e=IB/eaU_H=1.1·10^-29/m³=1.1·10^-23/cm³

39. B=μ₀IR²/2(L²+R²)^3/2=7.15·10^-6T, 40. B=μ₀e²/[8πr₀²(πε₀r₀m)^1/2]=12.5T

41. a) x=I₁d/(I₁+I₂)=0,6cm, między drutami, b) x=I₁d/(I₂-I₁)=3cm, po stronie I₁

Rozwiązania zadań z zestawu 5EM

42. ΔI₂=ΔI ^.R/πd=ΔI^.I₂/(I₁+2ΔI)=0,5A, 43. F=|F₁-F₂|=μ_oI₁I₂ab/2πd(d+b)=2.4·10^-5N

44. M=2F_m(b/2)sinφ, F_m=BIa, M=BIabsinφ, ab=S, Jε=-M, ε=φ”, φ≈0, sinφ≅φ,

φ”+(BIS/J)=0, ω²=BIS/J, T=2π/ω=2π(J/BIS)^1/2=0.63s

45. Φ_Bmax=_m/ω=_m/2πf=0.019Wb, 46. I=(E±BDv)/R=0,4A ∨ 0,2A

47. I₂(t₁)/I₁(t₁)=(a+b)²B_ofcos(2πft₁)/[2abA(t₁-t_o)]=47.35, 48. T=RCln[(-U_g)/(-U_z)]=6.73s

49. L=[(U₂/I₂)²-(U₁/I₁)²]^1/2/2πν=0.1H, 50. ε=[(2πνCU/I₁I₂)²(I₁²-I₂²)+1]^1/2=10.1

51. P=U_o²R/2[R²+(2πνL)²]=7.58W

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