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Conservation of charge

  1. Apr 6, 2005 #1
    The law of conservation of charge states that the net current due to flow of charge emanating from a closed surface S is equal to the time rate of decrease of the charge within the volume V bounded by S


    Can the time rate of increase of the charge in the cubical volume bounded by x,y,z planes be found?
  2. jcsd
  3. Apr 6, 2005 #2


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    What do you mean? Of course this applies to any volume, so I don't understand what exactly you're asking.
  4. Apr 6, 2005 #3
    when J = -2xi [A/m^2] I have to find the time rate in increase of charge in cubical volume bounded by planes x=0,x=2,y=0,y=2,z=0,z=2. I would like get hint to do this problem. Thanks
    Last edited: Apr 7, 2005
  5. Apr 6, 2005 #4


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    Current only passes through the x=0 and x=2 planes, so integrate over these. First find dS, the vector passing out perpendicular to the box with length dA, and the dot this with the current density, and integrate over each surface.
  6. Apr 7, 2005 #5
    So I will have?

    [tex]\frac{d}{dt}\oint_{V}\rho\*d\upsilon = -\oint_{S} -2x \cdot 12xdydz\vec{i}[/tex]

    [tex]\frac{dQ}{dt}= -\int_{0}^{2}\int_{0}^{2} -2x\vec i\cdot 12xdydz\vec{i}=96x^{2}[/tex]

    Is this correct?
    Last edited: Apr 7, 2005
  7. Apr 7, 2005 #6

    Your equation says a scalar equals a vector. Is that possible?
  8. Apr 7, 2005 #7
    [tex]\frac{dQ}{dt}= -\int_{0}^{2}\int_{0}^{2} -2x\vec i\cdot 12xdydz\vec{i}=96x^{2}[/tex]

    I have solved more difficult problems but I cannot do this one. I'm confused from the beginnig.
    Last edited: Apr 7, 2005
  9. Apr 7, 2005 #8


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    Where did that 12x come from? If you drop that and do the integral again, you will get the charging flow in the positive x direction over a 2X2 sheet in a plane parrellel to the x=0 plane. You want to find the total current leaving the box, so you need subtract the current going in at x=0 from the current leaving at x=2.
  10. Apr 7, 2005 #9
    [tex]\frac{dQ}{dt}= -\int_{0}^{2}\int_{0}^{2} -2x\vec i\cdot dydz\vec{i}=8x[/tex]

    So at x=2 current would be 16x?
  11. Apr 7, 2005 #10
    robert said: "I have solved more difficult problems...."

    I'm sure you have, because this one is VERY easy; you're making a lot harder than it really is.

    In what direction is the current density?

    Now define a vector that points out of the cube and in the direction that's perpendicular to the cube face that's at x=0.

    What direction is that vector pointing?

    If you did this for the other five faces (call them x=2, y=0, y=2....) what direction will each of your vectors be pointing?
  12. Apr 7, 2005 #11
    the current density is negative x direction
    The vector j points out of the cube and it is in the -y direction
  13. Apr 7, 2005 #12

    I have to apologize here. I missed what you said in your post #9. I saw your answer, 16x and thought you were pretty confused. But really you were very close. On the line above (where your integral is) you get dQ/dt = 8x. But then you evaluated that at x=2 and said that it was 16x????

    8x evaluated at x=2 isn't 16x. What is it?
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