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ACHE

  • 2 years ago

THREE CHARGES ARE AT THE corners of equilateral triangle.calculate electric feild at left corner charge?

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  1. henpen
    • 2 years ago
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    |dw:1349626330386:dw|

  2. henpen
    • 2 years ago
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    |dw:1349626433674:dw|

  3. henpen
    • 2 years ago
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    How would you actually do this? Well, define E2 to be on the x-axis, so \[E_{2(x)}=E_2\]\[E_{2(y)}=0\] And\[E_{1(x)}=E_1\cos(60)\]\[E_{1(y)}=E_1\sin(60)\]Add the x and y components

  4. henpen
    • 2 years ago
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    \[E=\frac{F}{q}=\frac{Qk}{r^2}\]

  5. ACHE
    • 2 years ago
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    i dnt understand:/

  6. henpen
    • 2 years ago
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    http://www.youtube.com/watch?v=OsWDUqJQcpk Watch the first 9 minutes, he explains something very similar to your problem on the 7th-9th minutes.

  7. ACHE
    • 2 years ago
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    utube iz bloked now:/

  8. ACHE
    • 2 years ago
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    theree charges are at the corners of an equilateral triangle as shown in fig below , calculte the electric feild at the position of tha 2.00 micro colum charge due to the 7.00 micro columb and -4.00 micro columb charges.(b) use ur answer to detrmine the force on the 2.00 m|dw:1349627210442:dw|icro columb chare.

  9. henpen
    • 2 years ago
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    |dw:1349627494021:dw| Are you fine this far?

  10. ACHE
    • 2 years ago
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    calculationx?

  11. henpen
    • 2 years ago
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    http://gyazo.com/01feb6fa59915fa730dc10d234ac6a87

  12. henpen
    • 2 years ago
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    +ve direction=right E1x=\[\frac{-4* (9*10^9)}{r^2}\] E2x=\[\frac{7* (9*10^9)}{r^2} cos(60)\] E2y=\[\frac{7* (9*10^9)}{r^2} sin(60)\]

  13. ACHE
    • 2 years ago
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    force?

  14. henpen
    • 2 years ago
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    \[E=F/q\]Multiply by the charge Sorry! In my earlier calculations it should have been MICRO coulombs- multiply each by 10^-6!

  15. henpen
    • 2 years ago
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    *Multiply E field by the charge to get force

  16. ACHE
    • 2 years ago
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    |dw:1349627940207:dw|

  17. ACHE
    • 2 years ago
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    r sq is o.5 sq. na?

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