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Integral of solids with linearly decreasing charge density

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the total charge embodied in a solid with charge density that decreases linearly with height from a value of λ at the bottom to 0 at the top.
    Solve for a rectangular prism and a sphere.

    2. Relevant equations

    ∫∫∫ρdxdydz
    ∫∫∫pr^2sinθdrdθd∅

    3. The attempt at a solution

    For a rectangular prism, I integrated from 0 to x,y for dx,dy and lambda to 0 for dz. I ended up with Q=x*y*ρ*some function of λ. However, something seems off about this approach, because the charge at any x and y varies with height.

    I think if I can get the logic behind the prism, I'll be able to understand the sphere, but any other advice would be helpful.
     
    Last edited: Jan 17, 2012
  2. jcsd
  3. Jan 17, 2012 #2
    Anyone have an idea?
     
  4. Jan 17, 2012 #3

    Curious3141

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    Homework Helper

    The charge enclosed in an element of height dx is given by [itex]\rho(x)A(x)dx[/itex] where [itex]\rho(x)[/itex] is a function that expresses the volume-charge density at a given height x from the base. A(x) denotes the area function, i.e. the cross-sectional area of a slice taken at height x from the base. x varies from 0 to h, the height of the figure.

    Work out an expression for [itex]\rho(x)[/itex] in terms of [itex]\lambda[/itex], h and x first (easy, since it's a linear relationship) - this applies to both figures. Then it's just a matter of evaluating the integral in each case between the bounds of 0 and h. For the rectangular prism (which is a cuboid), A(x) is a constant. For the sphere, A(x) follows the area of a disc. (For the sphere, h = 2R, where R is the radius).

    After you get the final expression, you should be able to rearrange to express it in terms of [itex]\lambda[/itex] and the volume V alone. When you do this, you will discover an interesting insight.
     
    Last edited: Jan 17, 2012
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