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anonymous

  • 4 years ago

the electric field strength 5cm from a very long charged wire is 2000N/C. what is the electric field strength 10cm from the wire?

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  1. Shayaan_Mustafa
    • 4 years ago
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    Use equation, E=F/q

  2. anonymous
    • 4 years ago
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    what i have is \[E=1/(4\pi \epsilon _{0}) * Q/r^{2}\]

  3. Shayaan_Mustafa
    • 4 years ago
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    but you don't have charge. so you can't use this relation.

  4. anonymous
    • 4 years ago
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    \[Q=L \lambda\] and the prof said use 2 for L. not sure why, maybe 10cm/5cm but why do that?

  5. anonymous
    • 4 years ago
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    after i solve for 5cm, i use the value i got for charge density to plug in for 10cm and got 500 N/C. i don't understand why L=2 and if i still use L=2 to solve for 10cm

  6. Shayaan_Mustafa
    • 4 years ago
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    there are two equations, 1) E=kq/r² find charge from above equation and put it in, 2) E=F/q find Force from above relation

  7. Shayaan_Mustafa
    • 4 years ago
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    did you got?

  8. Shayaan_Mustafa
    • 4 years ago
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    why L=2?? may be due to doubling the length.

  9. anonymous
    • 4 years ago
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    not really, i can't recall how to solve for force. force of field on wire or vice versa?

  10. anonymous
    • 4 years ago
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    or are you saying solve for force after i find E?

  11. Shayaan_Mustafa
    • 4 years ago
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    ok follow me.

  12. anonymous
    • 4 years ago
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    alright

  13. Shayaan_Mustafa
    • 4 years ago
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    use 1 equation and find Q by substituting values, k=9*10^9 E=2000 r=0.05m

  14. anonymous
    • 4 years ago
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    right

  15. anonymous
    • 4 years ago
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    Q=5.56*10^-10

  16. Shayaan_Mustafa
    • 4 years ago
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    ok good. Now use equation 2 and find F. by substituting values of, E=2000 q=5.56*10^-10

  17. anonymous
    • 4 years ago
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    should be 1.11*10^-6

  18. anonymous
    • 4 years ago
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    i will try to work more on it you will get a notification if i add more to this later, gotta go, thank you

  19. Shayaan_Mustafa
    • 4 years ago
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    you are welcome.

  20. JamesJ
    • 4 years ago
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    The electric field about a long, straight charged wire is given by \[ E(r) = \frac{\lambda}{2\pi \epsilon_0 r} \] where \( \lambda \) is the charge density. (See for example here: http://farside.ph.utexas.edu/teaching/302l/lectures/node26.html ) Given that, and what you are told about \( E(r) \) for \( r = 0.05 m \), you can then find \( E(r) \) for \( r = 0.10 m \).

  21. Shayaan_Mustafa
    • 4 years ago
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    hey jamesj who is this person who is reviewing 3rd question of emcrazy14?

  22. anonymous
    • 4 years ago
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    i am still confused. i solved for the charge density given 2000N/C for 5cm but i used 2 for L. L isn't given so i really don't understand why 2. well, it's the same wire the question is asking about so i used it and my answer is 500N/C which seems reasonable...

  23. JamesJ
    • 4 years ago
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    From the formula I wrote above for the electric field, you see that it scales as 1/r. So at twice the distance away it should be 1/2, not 1/4

  24. anonymous
    • 4 years ago
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    thx, i had my r squared, should be able to get it now

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