shows, in cross section, a plastic, spherical shell with uniform charge `Q =-16e` and radius `R = 10cm.` A particle with charge `q = + 5e` is at the center. What is the electric field (magnitude and direction) at (a) point `P _(1)` at radial distance `r _(1) = 6.00cm` and (b) point `P _(2)` at radial distance `r _(2) =12.0cm` ?

Explanation

To find the field at point `P _(1),` we consturuct a Gaussian sphere with `P _(1)` on its surface and thus with a radius of `r _(1).` Because the charge enclosed by the Gaussian sphere is positive, the electric flux through the surface must be positive and thus outward. So, the electric field `vecE` pierces the surface outward and, because of the spherical symmetery, must be radially outward, as drawn in Theat figure does not include the plastic shell because the shell is not enclosed by the Gaussian sphere. Consider a patch element on the sphere at `P _(1).` Its area vector `dvecA` is radially outward (it must always be outward from a Gaussian surface). Thus the angle `theta` between

(a) A charged plastic spherical shell encloses a charged particle. (b) To find the electric field at `P _(1),` arrange for the point to be on a Gaussian aphere. The electric field pierces outward. The area vector for the patch element is out ward. (c ) `P _(2)` is on a Gaussian sphere, `vecE` is inward, and `dvecA` is still outward.
Know that the surface area of a sphere is `4pir ^(2).` Substituting these results, for Gauss. law gives us
`epsi _(0) E 4 pi r ^(2) = q _(enc).`
The only charge enclosed by the Gaussian surface through `P _(1)` is that of the particle. Solving for E and substituting `q _(enc) = 5e and r = r _(1) = 6.00 xx 10 ^(-2)m,` we find that the magnitude of the electric field at `P _(1)` is
`=E= (q _(cnc))/(4 pi epsi _(0) r ^(2))`
`= ( 5 (1. 60 xx 10 ^(-19)C))/( 4pi ( 8.85 xx 10 ^(-12) C ^(2) //N.m^(2))(0.00600m)^(2))`
`=2.00 xx 10 ^(-6) N//C.`