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Gausses Law

  1. Sep 7, 2010 #1
    A point charge q1 = -7.5 μC is located at the center of a thick conducting shell of inner radius a = 2.9 cm and outer radius b = 4.9 cm, The conducting shell has a net charge of q2 = 2.3 μC. (see attachment)


    1. What is Ex(P), the value of the x-component of the electric field at point P, located a distance 7.5 cm along the x-axis from q1?

    2. What is Ey(R), the value of the y-component of the electric field at point R, located a distance 1.45 cm along the y-axis from q1?

    3. What is σb, the surface charge density at the outer edge of the shell? C/m2

    4. What is σa, the surface charge density at the inner edge of the shell? C/m2

    Ok so I am not very familier with the gauss law and how I set up the equations and such.
    Lets just start with number 1.
    I know I have to use the law: E= Q/ (4 *pi* eo *r)
    where Q is the charge of q1 and r=.029m. I just dont know how the net charge q2 and its radius plays a role into the equation (i know it does but i just dont understand where to plug that in). is Q the difference of q1 and q2? how about r?

    Number 3- Surface charge density is E*eo.
    So to find this out, I just use E= Q/ (4 *pi* eo *r) where Q=q1 and r=.029 and eo is 8.85*10^-12


    Any help would be great. THANKS!
     

    Attached Files:

  2. jcsd
  3. Sep 8, 2010 #2

    collinsmark

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    Hello mattmannmf,

    Let me start by carrying something over to this thread that I just posted yesterday on ta different thread:

    Gauss' law states that

    [tex] \oint _S \vec E \cdot \vec{dA}= \frac{Q_{ENC}}{\epsilon _0} [/tex]

    What this means is that the dot product of the electric field and the differential area vector, summed over all area in a closed surface (any arbitrary close surface), is proportional to the total charge enclosed within that surface. More on how to use this idea below.

    Gauss's law is pretty neat, and it can be used all the time for situations. But in the case where you are using it to find the electric field, it's really only useful in a handful of specific situations, and only where symmetry is involved. As a matter of fact, it's really only useful in finding the electric field if you dealing spherical points, shells or spheres (things with spherically symmetrical charge distribution); infinitely long (or at least very long) line charges or cylinder charges (things with cylindrical symmetry), or infinitely long (or at least very long) planes (things with planar charge distribution).

    The key to using to find the electric field is to follow these guidelines:
    (1) Pick a Gaussian surface such that the electric field is perpendicular to the surface (i.e. parallel to the normal vector of the surface) and the magnitude of the electric field is constant all across the entire surface. (You can ignore the caps of cylinders [and/or sides of 'pillboxes'] because the contribution is negligible when considering infinitely cylinders and planes, and secondly because the electric flux through the caps of a Gaussian cylinder is zero [dot product of the vectors is zero] for the caps anyway, when considering infinitely long line and cylinder charges.)
    (2) If you've followed the first step, the integral is a snap, because [itex] \oint E \cdot dA = EA [/itex] if the conditions in guideline 1) are met.
    (3) Solve for the magnitude of E.

    All that being said, Guass' law is always useful in the respect that if you already happen to know what the electric field is, you can always use Guass' law to find the charge enclosed inside of a hypothetical closed surface.
    Umm, not quite the right approach. :uhh:

    Consider a hypothetical, spherical surface centered around point q1 (centered around the origin). Suppose the radius of this spherical surface is 7.5 cm, such that the surface passes through point P. Ask yourself the following questions:
    (i) How much total charge is inside this hypothetical surface?
    (ii) What is the surface area of this hypothetical surface (i.e. what is the surface area of a sphere of radius 7.5 cm)?

    Once you answer (i) and (ii) you have enough information to solve part 1.

    This hypothetical surface that we are using is called a "Gaussian surface".

    Follow the same sort of idea for part 2, but now the thick shell is outside the hypothetical, spherical, surface. So how much of the thick shell, thus how much of the thick shell's charge, is inside the hypothetical spherical surface? (Hint: it ain't much.) But the point charge q1 is inside the surface. So what's the total charge inside the Gaussian surface? What's the surface area of the Gaussian surface in this case?

    Again, I don't think that's quite the right approach.

    Parts 3 and 4, although no more complicated, are a little trickier. They rely on two key characteristics regarding conductors:
    (a) The electric [electrostatic] field within a conducting material is 0.
    (b) Any residual charge is only located at the surfaces of the conductor. The surface charge distribution aligns itself such that the electrostatic field within the conducting material itself is zero.

    With that in mind, if you change the radius of your Gaussian (hypothetical) surface such that the Gaussian surface is inside the very material of the thick shell (a < radius < b), what must the charge be on the inside surface of the thick, conducting shell (Hint: since the electric field is zero, you know that the total charge inside the Gaussian surface must be zero -- so what must be the charge be on the inside of the thick, conducting shell, given that there is a point charge q1 in there too)? Since you already know the total charge of the conducting shell, and you've just figured out the inside surface charge on the conducting shell, what must be the charge on the outside surface of the thick, conducting shell?

    [Edit: Of course, once you find the surface charges on the inside and outside surfaces of the thick, conducting shell, you will have to divide by the respective surface areas to obtain the respective charge densities.]
     
    Last edited: Sep 8, 2010
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