Electric field due to an infinite line of charge

Electric field due to an infinite line of charge.

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  Since Coulomb's law can only be applied to point charges, we could start by dividing the given line into infinitesimal pieces, making each piece, a point charge.

  Okay! Then what?

  We could then add the electric field due to each piece, at point $P$‍, giving us the total field.

  But there are infinitely many pieces!

  Yep, and so this would involve an integral. Yuck! The whole point of using Gauss's law is to avoid this integral.

Simplifying the left hand side of Gauss's law

• Top surface

  Let's start with the top surface

  ${\int }_{top}\overrightarrow{E}.\overrightarrow{dA}=$‍

  • Your answer should be
  • an integer, like $6$‍
  • a simplified proper fraction, like $3/5$‍
  • a simplified improper fraction, like $7/4$‍
  • a mixed number, like $13/4$‍
  • an exact decimal, like $0.75$‍
  • a multiple of pi, like $12\text{pi}$‍ or $2/3\text{pi}$‍

  CheckExplain
• Bottom surface

  Now let's try the bottom surface

  ${\int }_{bottom}\overrightarrow{E}.\overrightarrow{dA}=$‍

  • Your answer should be
  • an integer, like $6$‍
  • a simplified proper fraction, like $3/5$‍
  • a simplified improper fraction, like $7/4$‍
  • a mixed number, like $13/4$‍
  • an exact decimal, like $0.75$‍
  • a multiple of pi, like $12\text{pi}$‍ or $2/3\text{pi}$‍

  CheckExplain

Flux through the curved surface

• Curved surface

  What is the flux through the curved surface, ${\int }_{curved}\overrightarrow{E}.\overrightarrow{dA}=$‍?

  Choose 1 answer:

  Choose 1 answer:

  • (Choice A)  

    0

    A

    0
  • (Choice B)  

    $E.A$‍

    B

    $E.A$‍
  • (Choice C)  

    $E.(\pi r^2)$‍

    C

    $E.(\pi r^2)$‍
  • (Choice D)  

    $E.(2\pi rL)$‍

    D

    $E.(2\pi rL)$‍
  • (Choice E)  

    $E.(\pi r^{2}L)$‍

    E

    $E.(\pi r^{2}L)$‍

  Check

Simplifying the right-hand side of Gauss law.

Simplifying and finding the electric field strength.

Summary & more on cylindrical symmetry