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Electric field in the overlap of two solid, uniformly charged spheres

  1. Oct 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Two spheres, each of radius R and carrying uniform charge densities +[tex]\rho[/tex]
    and [tex]-\rho[/tex], respectively, are placed so that they partially overlap.
    Call the vector from the positive centre to the negative centre [tex]\vec{d}[/tex]. Show
    that the field in the region of overlap is constant and find its value. Use
    Gauss’s law to find the electric field inside a uniformly charged sphere
    first.


    2. Relevant equations
    law of superposition
    Gauss Law

    3. The attempt at a solution
    I found the field inside one sphere to be
    [tex](r\rho)/(3\epsilon)[/tex]
    in the radial direction. Now for the overlapping spheres, I said that the vector from the centre of the positive sphere to some point P in the interlapping area is [tex]\vec{r}[/tex]. And from P to the centre of the negative sphere, I denoted [tex]\vec{r'}[/tex]. so [tex]\vec{r'}=\vec{d}-\vec{r}[/tex]. So in order for P to be inside the spheres, [tex]|\vec{r}|<R[/tex] and [tex]|\vec{d}-\vec{r}|<R[/tex]. So using the law of superposition, inside the overlap, the electric is
    [tex]E = (|\vec{r}|-|\vec{d}-\vec{r}|)\rho/3\epsilon[/tex]
    in the radial direction, with the boundaries in effect. Now I am stumped here, as I'm unsure how to reduce this to a constant. Any suggestions?
     
  2. jcsd
  3. Oct 6, 2010 #2
    the electric field is a vector so why on earth are you reducing r and d-r to scalars?
     
  4. Oct 7, 2010 #3
    yep, realized my mistake while sitting in my thermodynamics class this morning. It's funny how I usually figure stuff out while I'm not actually trying to do the question.
     
  5. Oct 7, 2010 #4
    k, so I changed the E function to Cartesian coordinates. So in the overlap I got:

    [tex]\vec{E}=\frac{\rho*d}{3*\epsilon}*\hat{i}[/tex]

    where d is the magnitude of [tex]\vec{d}[/tex]

    And this is under the condition that [tex]\vec{d}[/tex] runs along the x axis.
     
  6. Oct 7, 2010 #5
    sometimes you just need to sleep on it and get a fresh perpective on it in the morning
     
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