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Do you integrate a variable charge density when solving for Q?

  1. Nov 29, 2013 #1
    For example, let's say there was an block infinitely long in the x and y direction and in the z direction bounded by positive and negative a.

    I am trying to find the charge of a imaginary partition infinitely long in the x and y direction and in the z direction bounded by positive and negative b, where b is less than a.

    Here is where I get confused, if the charge density is something like p = p0*(a^2 - z^2), where z in the case is b, as mentioned above.

    dQ = p * dV

    Do i integrate from here or do I substitute in p = p0*(a^2 - z^2), and then integrate?
  2. jcsd
  3. Nov 29, 2013 #2


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    As p is equal to p0*(a^2 - z^2) it does not matter, but I wonder how you try to integrate it without replacing p by the other expression.
  4. Nov 29, 2013 #3
    I thought to just take p out of the integral and substitute it in later after I integrate dV with respect to z. Also, p should still matter because I stated that z^2 is in this case b^2.
  5. Nov 29, 2013 #4

    Simon Bridge

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    Perhaps this will help: ##\renewcommand{\d}{\text{d}}##
    The charge in the infinite sheet between ##z## and ##z+dz## is $$\d q=p(z)\d z$$
    So the charge within ##|z|<b## is the sum of all such sheets inside that range.

    In your example, "##z^2=b^2##", as you say, only at the point ##z=b## ... but the total charge includes all the charge at all points between z=-b and z=b as well - so z has to be able to take on all values in between. vis: $$q=\int_{-b}^b p(z)\;\d z$$ ... i.e. you cannot take ##p## outside the integral because it depends on the variable being integrated over.

    There may be a small confusion here though.
    $$\d q=\rho(x,y,z) \d V = p(z)\d z : p(z)=\int_{-\infty}^\infty\int_{-\infty}^\infty\rho(x,y,z)\; \d x\d y$$ ... the provided density function has already done two of the integrations for you - so you are not integrating dV over z.
    Last edited: Nov 29, 2013
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