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Gina

  • 5 years ago

help me please!!,,,to the plate plane air capacitor applied potential difference U = 500V in the area of ​​the plates with a = 200 cm square, the distance between d1 = 1,5 mm. plates moved apart to a distance d2 = 15mm. find the energy E1 and E2 of the capacitor before and after razdvizheniya plates, if the source of EMF to razdvizheniem: 1) switched off, and 2) does not turn off. (1) E1 = 14,8 mJ, E2 = 148mkdzh; 2) E1 = 14,8 mJ: E2 = 1,48 mJ

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  1. sgadi
    • 5 years ago
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    can you explain what is "razdvizheniem"

  2. Gina
    • 5 years ago
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    that is supposed to the word( motion),,,i translated from russian to english

  3. sgadi
    • 5 years ago
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    you have given answer also ... you need procedure ... right ?

  4. Gina
    • 5 years ago
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    yes,plesae

  5. sgadi
    • 5 years ago
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    answers are in milli J or in micro J .... coz ... I am getting answers in micro J

  6. Gina
    • 5 years ago
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    \[\int\limits_{o}^{a}d _{x}\]

  7. Gina
    • 5 years ago
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    what is important is the procedure os can you show me ?

  8. sgadi
    • 5 years ago
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    I am typing ... I will send in few minutes ..

  9. Gina
    • 5 years ago
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    o right

  10. sgadi
    • 5 years ago
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    Formulas required \[C=\epsilon_0\frac{A}{d}\] \[Q=CV\] \[E=\frac{1}{2}CV^2\] given \[A=200 cm^2 = 0.02 m^2\] \[d_1=1.5 mm= 1.5e-3 m\] \[d_2=15 mm= 15e-3 m\] Calculating Capacitance: \[C_1=1.1805e-010\] \[C_2=1.1805e-011\] Case 1: when external supply voltage is cut off before movement voltage is give as \[V_1=500\] Finding charge \[Q=C_1V_1=5.9027e-008\] after movement charge is constant. So voltage changes to \[V_2=\frac{Q}{C_2}=5000\] Finding energy \[E_1=\frac{1}{2}C_1V_1^2=1.4757e-005 \approx 14.8\mu J\] \[E_2=\frac{1}{2}C_1V_1^2=1.4757e-004 \approx 148 \mu J\] Case 2: when external supply voltage is maintained at 500V Voltage is constant V_1=V_2=V=500v Finding energy: \[E_1=\frac{1}{2}C_1V^2=1.4757e-005 \approx 14.8\mu J\] \[E_2=\frac{1}{2}C_1V^2=1.4757e-006 \approx 1.48\mu J\]

  11. sgadi
    • 5 years ago
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    Gina .... I did a mistake .... in the formula of E2 .... the capacitance is not C1 ... it is C2

  12. Gina
    • 5 years ago
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    wow!! thank you very much))

  13. sgadi
    • 5 years ago
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    welcome

  14. Gina
    • 5 years ago
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    ok i will check on it

  15. Gina
    • 5 years ago
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    mr.sgadi do have the notes on this topic?

  16. sgadi
    • 5 years ago
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    I don't have any special notes regarding this topic. I used Wikipedea. http://en.wikipedia.org/wiki/Capacitance

  17. Gina
    • 5 years ago
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    thanx again)

  18. sgadi
    • 5 years ago
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    :-) welcome

  19. Gina
    • 5 years ago
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    what this about this question of maths,,,,∫oadx

  20. sgadi
    • 5 years ago
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    I didn't understand. Can you elaborate?

  21. Gina
    • 5 years ago
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    \[\int\limits_{0}^{a?}d _{x}\]

  22. Gina
    • 5 years ago
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    can help mathamatics?

  23. sgadi
    • 5 years ago
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    yes... to an extent. place in new thread so that everyone will be able to help you.

  24. Gina
    • 5 years ago
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    can help mathamatics?

  25. sgadi
    • 5 years ago
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    yes. please give the question

  26. Gina
    • 5 years ago
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    \[d _{x}\div \sqrt{x ^{2}+x+1}\]

  27. sgadi
    • 5 years ago
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    \[\int{\frac{1}{\sqrt{x^2+x+1}}dx}=\ln{(x+\sqrt{x^2+x+1}+0.5)}\]

  28. Gina
    • 5 years ago
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    yes that is the question

  29. Gina
    • 5 years ago
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    is that the solution?

  30. sgadi
    • 5 years ago
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    yes ... I have given the solution also

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