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woleraymond

  • 2 years ago

An infinite charged sheet has a surface charge density of 1 x 10^-7 Cm^-2 .How far apart are the equipotential surfaces whose potentials differ by 5volts?

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  1. henpen
    • 2 years ago
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    Work out the electric field at any point, and use the fact that potential energy is \[-\int\limits_{\infty}^{a}F(x)dx\](i.e. a unit charge is dragged from infinity, parallel to the field lines, to a)

  2. henpen
    • 2 years ago
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    Use the fact that the electric field is the same everywhere (think of the electric flux through a cylinder on the plane).

  3. Mikael
    • 2 years ago
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    Simpler when the field is constant which it is here - always for infinite plane \[\Delta V = E*\Delta x\]

  4. Mikael
    • 2 years ago
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    so 5 = E*Delta x find E and then it will help find\[\Delta x\]

  5. henpen
    • 2 years ago
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    Yes, basically my integral simplifies to \[U=-Fx\] As F is not actually a function of x. Evaluated from infinity to a, or more simply the change in PE (from b to a), yields Mikael's result.

  6. RaphaelFilgueiras
    • 2 years ago
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    |dw:1347905879599:dw|

  7. henpen
    • 2 years ago
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    http://www.youtube.com/watch?v=ldJhMDuOGxY

  8. woleraymond
    • 2 years ago
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    @RaphaelFilgueiras pls can you be more legible with ur formulas.....@benpen thanks for the link

  9. woleraymond
    • 2 years ago
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    but the value i get is 8.98 x 10 ^ -4 m @RaphaelFilgueiras any clues

  10. RaphaelFilgueiras
    • 2 years ago
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    it's close my is 8.85x10^-4

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