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GR0877 Problem 62

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) Gauss’s law: \Phi_{total} = \Phi_A + \Phi = q/\epsilon_0; \Phi = q/\epsilon_0 - \Phi_A = \displaystyle   = \frac{1 \cdot 10^{-9}}{8.85 \cdot 10^{-12}} + 100 \approx 100 + 100 = 200 (\text{N} \cdot \text{m}^2/C). Check the “table of information” in your test book for the numeric value of \epsilon_0 (if you don’t remember).

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Categories: 0877 Tags: , , , ,
  1. George
    08.11.2011 at 03:29

    They give that the flux is negative 100, thus to achieve the total flux of 100 we know must be exiting the entire container, the rest of the surface must emit 200 units of flux. i.e. (unknown flux through rest of container) -100 = 100

  2. Pedro
    07.11.2012 at 18:53

    So this is basically saying that the -100 flux through the opening A is due the to same charge that is inside? That doesn’t make sense as the charge is positive and the flux through A negative. If the source of the flux through A is outside the container, than the equation in this solution above isn’t correct.

    • giraffe
      23.10.2014 at 14:57

      It says “net charge” is positive, and perhaps the distribution of the total charge inside the cylinder is not uniform. I mean like here are more negative charges and there are more positive charges.

  3. 11.11.2012 at 21:10

    Pedro :

    So this is basically saying that the -100 flux through the opening A is due the to same charge that is inside?

    NO. This is saying what mister Johann Carl Friedrich Gauss was saying: The electric flux through any closed surface is proportional to the enclosed electric charge.

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