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anonymous

  • 4 years ago

two 10cm diameter charged rings face each other 20cm apart. both rings are charged to +40nC. what is the electric field strength...

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  1. anonymous
    • 4 years ago
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    a) at the midpoint between 2 rings? should the E just be added together but using r=10cm?

  2. EarthCitizen
    • 4 years ago
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    \[E=F/Q\]

  3. anonymous
    • 4 years ago
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    found the answer on cramster step by step but not understanding the r: Electric field due to charged ring E = KQx / [ x ^ 2 + R^2 ] ^ 1.5

  4. anonymous
    • 4 years ago
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    (x^2+R^2)^1.3??

  5. anonymous
    • 4 years ago
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    oops 1.5

  6. EarthCitizen
    • 4 years ago
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    is it between to plates ?

  7. anonymous
    • 4 years ago
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    yes

  8. EarthCitizen
    • 4 years ago
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    two plates*

  9. anonymous
    • 4 years ago
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    yes

  10. EarthCitizen
    • 4 years ago
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    \[\epsilon _{0}\ = ?]

  11. anonymous
    • 4 years ago
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    8.85*10^-12

  12. anonymous
    • 4 years ago
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    k=1/(4pi8.85*10^-12)=8.99*10^9

  13. EarthCitizen
    • 4 years ago
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    first convert cm to m

  14. anonymous
    • 4 years ago
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    right

  15. anonymous
    • 4 years ago
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    .1m diamter, .2m apart

  16. anonymous
    • 4 years ago
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    and +40nC=+

  17. anonymous
    • 4 years ago
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    +40*10^-9

  18. anonymous
    • 4 years ago
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    answer says it should be 0 cuz they cancel each other but i thought it would double since both rings are +

  19. EarthCitizen
    • 4 years ago
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    both rings are either + /-

  20. EarthCitizen
    • 4 years ago
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    calculate the force, r= d/2

  21. anonymous
    • 4 years ago
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    i actually understand now from the set up on cramster except for the part that is raised to 1.5...typo? should be .5

  22. anonymous
    • 4 years ago
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    o charge density for ring=Q/(x^2+r^2)^1.5

  23. anonymous
    • 4 years ago
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    i mean Q/(2pi*r)

  24. EarthCitizen
    • 4 years ago
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    r=d/2

  25. EarthCitizen
    • 4 years ago
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    d=0.1m/2 = 0.05m

  26. EarthCitizen
    • 4 years ago
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    total net E=E1+E2

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