In Homework Problem 2.1, you were given the spherically symmetric charge density

Question

In Homework Problem 2.1, you were given the spherically symmetric charge density rho(r) = {rho_0 r^2/R^2 r R from which you calculated the field E(r) = rho/5elementof_0 {r^3/R^2 r r R (a) Using the relationship given in Problem 3.1(a), calculate the potential V(r) under the assumption the reference point r_0 is somewhere at infinity. (b) Using the inverse: relationship E = -Delta V, check that the potential found in (a) is consistent with the given field. Answer: (a) V(r) = rho_0/20elementof_0 {(5R^2 - r^4/R^2) r R.

Explanation / Answer

Given,
for r < R
charge density, rho(r) = rhoo*r^2/R^2 [ where rhoo is charge density of the sphere, R is its radius)]
Electric field, E(r) = rhoo*r^3/5*epsilon*R^2
so using the formula for potential
V(r) = integrate(E(r)dr) from r to 0 [ assuming 0 potential at infinity , and V potential at r]
V(r) = integrate(rhoo*r^3/5*epsilon*R^2) dr
V(r) = rhoo*r^4/20*epsilon*R^2 + K [ where K is a constant]

for r >= R

charge density, rho(r) = 0 [ where rhoo is charge density of the sphere, R is its radius)]
Electric field, E(r) = rhoo*R^3/5*epsilon*r^2
so using the formula for potential
V(r) = integrate(E(r)dr) from infinity to r [ assuming 0 potential at infinity , and V potential at r]
V(r) = integrate( rhoo*R^3/5*epsilon*r^2) dr
V(r) = rhoo*R^3/5*epsilon*r

at the surface,
from the last two formulas V(R) = rhoo*R^4/20*epsilon*R^2 + K = rhoo*R^3/5*epsilon*R
K = rhoo/5*epsilon - rhoo/20*epsilon
K = 4*rhoo/20*epsilon

a.    so V(r) for r < R = rhoo(5R^2 - r^4/R^2)/20 epsilon
       V(r) for r > R = rhoo(4R^3/r)/20*epsilon
b. for r < R
   -d(V(r)/dr) = -d(rhoo(5R^2 - r^4/R^2)/20 epsilon)/dr = rhoo*r^3/5*epsilon*R^2 = E(r) for r < R
   for r > R
   -d(V(r)/dr) = -d(rhoo(4R^3/r)/20*epsilon)/dr = rhoo*R^3/5*epsilon*r^2 = E(r) for r > R

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