# Gauss' Law?

## Figure shows a point charge q = 126μC at the center of a spherical cavity of radius 3.66cm in a piece of metal. Use Gauss’ law to find the electric field (a) at point P1, halfway from the center to the surface, (b) at point P2.

Dec 13, 2017

#### Explanation:

The electric field inside a good conductor cannot be non-zero. If the electric field has to vanish inside a conductor, it must get polarized such that $- 126 \setminus \mu C$ of polar charge will get distributed at the cavity wall in such a way that the field lines due to the central charge will terminate on these polar charges.

But since the conductor is electrically neutral, each of these negative polar charges will have a positive charge of same magnitude associated with them. These would have drifted to the surface of the good conductor. The electric field will continue outside the conductor originating from these outer $+ 126 \setminus \mu C$.

Interior: Let $r$ be the distance of point ${P}_{1}$ from the central charge ($q = + 126 \setminus \mu C$). Construct an imaginary gaussian sphere centered on the central charge and of radius $\frac{r}{2}$, so that the point ${P}_{1}$ lies on its surface.

By Gauss's law the electric flux through this surface is -

\oint\quad vecE.dvecs = q_{"tot"}/\epsilon_0;

The electric field is spherically symmetric and so will have the same magnitude (${E}_{1}$) everywhere on the surface of this sphere. This integral then reduces to -

E_1 [4\pi(r/2)^2] = q_{"tot"}/\epsilon_0; \qquad E_1 = q/(\pi\epsilon_0r^2)

q=+126\muC; \qquad r=

Exterior: Point ${P}_{2}$ lies inside the good conductor. So the electric field at that point vanishes.

If we imagine a gaussian sphere centered on the central charge such that point ${P}_{2}$ lies on its surface, the total charge inside this sphere is zero. Because the total charge includes the central charge $q = + 126 \setminus \mu C$ and the polarized charge ${q}_{p}^{-} = - 126 \setminus \mu C$.

\oint\quad vecE.dvecs = q_{"tot"}/\epsilon_0; \qquad

E_2[4\pir_2^2] = (q+q_p^{-})/epsilon_0 = 0;
Therefore, $\setminus \quad {E}_{2} = 0$.