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Electrostatic force

  1. Feb 22, 2017 #1
    1. The problem statement, all variables and given/known data
    A uniformly charged, solid, non-conducting sphere of radius ##a## and charge density ##\rho## has its center located a distance ##d## from a uniformly charged, non-conducting, infinite sheet (charge density ##\sigma##). Determine the potential difference between the center of the sphere and the nearest point on the charged sheet. Determine the net electrostatic force on the charged sphere.

    2. Relevant equations
    ##E=-\nabla V##
    ##F=qE##
    3. The attempt at a solution
    I tried something but I want to make sure it is correct. So the electric field created by the plane is ##E_{plane}=\frac{\sigma}{2\epsilon}## and ##V_{plane}=\frac{\sigma r}{2\epsilon}##. Using Gauss law the potential inside and outside the sphere is: ##V_{sphere}^{inside}=\frac{2\rho r^2}{3\epsilon}## and ##V_{sphere}^{outside}=\frac{Q}{4 \pi \epsilon r}##. So at the center of the sphere the potential is the one created by the plane, so ##V_1 = \frac{\sigma d}{2\epsilon} ## and the potential at the surface of the plane is the one created by the sphere ##V_2 = \frac{Q}{4 \pi \epsilon d}##. So the potential difference would be just ##V_1 - V_2##?
    For the second part, the electric field of the plane is constant, so the force is just ##\frac{\sigma}{2\epsilon} Q##? With ##Q = \rho 4 \pi r^3/3## Thank you.
     
  2. jcsd
  3. Feb 22, 2017 #2

    TSny

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    Homework Helper
    Gold Member

    You haven't been consistent with where you are choosing V to be zero. For the plane, you are choosing V = 0 at the surface of the plane. For inside the sphere, you are choosing V = 0 at the center of the sphere, and for outside of the sphere, you are choosing V = 0 at infinity.

    Also, I'm not sure about the factor of 2 in the numerator of the expression for the potential for inside the sphere.

    You might consider getting ΔV by integrating the field between points 1 and 2.

    I believe your answer for the second part is correct.
     
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