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Electric potential and field of sphere.

  1. Oct 24, 2012 #1
    In a sphere of radius R the charge density is given by:

    p(r) = Q*r/(pi*R^4) , the r is the distance of a generic point to the center of sphere.


    a) Confirm that the total charge is equal to Q.

    b) What is the electric field inside and outside sphere.

    c) What is the electric potential inside and outside sphere when the zero potential is considered in infinite and the potential function is continuous in r=R.


    Attempts:

    a) dq = ∫p dV = ∫Q*r/(pi*R^4) dV= Q(pi*R^4)/∫rdV

    in spherical coordinates we obtain:

    Q/(pi*R^4)∫(0->2pi) ∫(0->pi)∫(0->r) r*r^2*sinθdrd[itex]\Phi[/itex]

    Q/(pi*R^4)*(4pi*r^4/4pi)

    to have the total charge in sphere r must be = R so

    Q/(pi*R^4)*(4pi*R^4/4) = Q


    b) Inside sphere r < R

    so the inner charge is = (Q/R^4)*(r^4)

    then by the electric flux

    E(4*pi*r^2) = (Q/R^4)*(r^4)*(1/εo) vector E have radial direction

    E = (Q/R^4)*(r^2)/(4i*εo) N/C


    Outside sphere r>R

    the charge is Q

    then by the electric flux

    E(4*pi*r^2) = Q/εo

    E = Q/(4*pi*r^2*εo) N/C


    c) The electric potential outside the sphere is :

    r>R

    Point 1 = arbitrary point outside sphere
    V1(r)-V(∞) = -∫(infinity to r) Q/(4*pi*r^2*εo) dr = Q/(4*pi*r*εo) V


    attempt 1:

    Point 2 = arbitrary point inside sphere
    V2(r)-V(∞) = -∫(infinity to R) Edr (r>R) - ∫(R to r)Edr (r<R) =
    = Q/(4*pi*R*εo) - ∫(R to r)(Q/R^4)*(r^2)/(4i*εo)dr .......


    Can somebody check if the solutions in a and b are correct?. And c, am i proceeding it correctly?
     
    Last edited: Oct 24, 2012
  2. jcsd
  3. Oct 24, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Looks good to me. Does the integration limit "a" stand for R in the last couple of lines?
     
  4. Oct 24, 2012 #3

    yes it does.
     
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