anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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)
anonymous
  • anonymous
Use the fact that the electric field is the same everywhere (think of the electric flux through a cylinder on the plane).
anonymous
  • anonymous
Simpler when the field is constant which it is here - always for infinite plane \[\Delta V = E*\Delta x\]

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anonymous
  • anonymous
so 5 = E*Delta x find E and then it will help find\[\Delta x\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
|dw:1347905879599:dw|
anonymous
  • anonymous
http://www.youtube.com/watch?v=ldJhMDuOGxY
anonymous
  • anonymous
@RaphaelFilgueiras pls can you be more legible with ur formulas.....@benpen thanks for the link
anonymous
  • anonymous
but the value i get is 8.98 x 10 ^ -4 m @RaphaelFilgueiras any clues
anonymous
  • anonymous
it's close my is 8.85x10^-4

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