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Homework Help: Electrical Physics

  1. Jul 12, 2006 #1
    Hey Everyone,
    I had the great idea to enroll in this physics as a distance ed. course. I quickly have discovered that I am supposed to understand everything just by reading the text. Anyhow I have come across some problems that whatever I do I can't figure out so here it goes:
    1. Flux and nonconducting shells.
    A charged particle is suspended at the center of two concentric spherical shells that are very thin and made of nonconducting material. Figure 23-31a shows a cross section. Figure -1-31b gives the net flux through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere.
    Now, the diagram illustrates that at the point of interest, the net flux is -4*10^5 and asks what the charge of ring "A" is. To solve for this I tried many things: 1. using the equation qenc=Flux*epsilon*naut (didn't work), Next I tried subtracting the central charge from the net Flux*epsilon*naut at the point of interest. Neither of these worked.

    2.Two charged concentric spherical shells have radii of 10.0 cm and 15.0 cm. The charge on the inner shell is 4.10*10-8 C and that on the outer shell is 2.50*10-8 C. Find the magnitude of the electric field at r=11.0 cm and r=20.5 cm.
    Honestly I didn't know where to start, so I just looked at the formula in the book E=8.99*10^9*(q/R^3)*r. Where q=4*10^-8, R=10cm and r=11cm.
    I really appreciate any help with this.
  2. jcsd
  3. Jul 12, 2006 #2

    Andrew Mason

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    Homework Helper

    We will need either Figure 1-31b or its contents in order to help you answer this question.

    I can see why you are confused. The answer given is not right.

    This appears to be just an application of Gauss' law or Coulomb's law, if we assume that the charge is evenly distributed over the area of the two shells.

    [tex]\int E\cdot dA = E*4\pi r^2 = q_{encl}/\epsilon_0[/tex]

    Or, by application of Coulomb's law:

    [tex]E = kq/r^2[/tex]

    For the first part,

    [tex]E = q_{1}/4\pi\epsilon_0r^2 = kq_1/r^2[/tex]

    For the second,

    [tex]E = (q_1 + q_2)/4\pi\epsilon_0r^2 = k(q_1+q_2)/r^2[/tex]

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