Its direction is along (or opposite to) the radius vector r
from the origin to the point P The obvious Gaussian surface is a
spherical surface centred at the nucleus We consider two situations,
namely, r < R and r > R (i) r < R : The electric flux f enclosed by the spherical surface is
    f  =  E (r) × 4 p r 2
*
Compare this with a uniform mass shell discussed in Section 7