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Electric field question!

  1. Feb 4, 2005 #1
    i am trying to calculate the enclosed charge but i am having trouble figuring out how to integrate E. E is said to be nonuniform and given by the equation E= (a + bx^2)i where a = 4 n/c and b = 2 n/c*m^2. What exactly do i integrate here? i am very confused. (ps. i have the surface area that the electric field is flowing through perpendicularly... any help would be greatly appreciated!!

  2. jcsd
  3. Feb 4, 2005 #2
    Your question is not clear.
    Explain the symbols, at least. What is x, what is the surface, what is i.
    Try to write things clearly.
    Eventually, write down the exact question in full, exactlt as you have been asked it.
  4. Feb 4, 2005 #3
    Welcome to PF! please take a second to read the sticky (its the first thread entitled "Read this before posting")
    please don't double post your problems...
    Last edited by a moderator: Feb 4, 2005
  5. Feb 4, 2005 #4

    Doc Al

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    Staff: Mentor

    Gauss's Law

    (I merged the two threads into one.)

    Here's a hint: You are looking to apply Gauss's law, which relates the electric flux through a closed surface with the charge enclosed. Look it up!

    As MathStudent advises, read the sticky. Show your work and you'll get plenty of help. Give it a shot.
  6. Feb 4, 2005 #5
    here's the problem....

    ok... the problem says that there is an enclosed surface with a=b=.436 m and c=.582 m. (forming a rectangular prism)

    the electric field is passing perpendicularly through the surface a*b which is .190 m^2

    it goes on to say that the electric field is non uniform and is given by the equation E=(alpha + beta*x^2)i where alpha = 4 N/C and beta =2 N/C*m^2

    I am trying to calculate the enclosed charge...

    I know that:
    flux = sum of E * delta A
    flux = charge/ epsilon0
    so sum of electric field * surface area it is going through = charge/epsilon0?

    charge = sum of electric field * surface area it is going through * epsilon0

    and to get the sum of the electric field i feel like i need to take the integral of that equation for E above with alpha and beta.... is this right? and i'm not exactly sure how to integrate this equation... help! and thank you guys so much
  7. Feb 4, 2005 #6

    Doc Al

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    Staff: Mentor

    To find the flux through that surface you need to "integrate" E*dA over the surface. But where is surface a*b? It's location is important, since the field varies with x coordinate.

    Hint: The only surfaces with non-zero flux are those that have some component of the field perpendicular to them. If I picture your closed surface correctly, you have two surfaces with non-zero flux, both having area a*b. But where are they?
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