Question

(figure 1) a spherical cavity is hollowed out of the interior of a neutral conducting sphere. at the center of the cavity is a point charge, of positive charge ( q ). part a: what is the total surface charge ( q_{\text{int}} ) on the interior surface of the conductor (i.e., on the wall of the cavity)? part b: submit box part c: what is the magnitude ( e_{\text{int}} ) of the electric field inside the cavity as a function of the distance ( r ) from the point charge? let ( k ), as usual, denote ( \frac{1}{4piepsilon_0} ). options: ( 0 ), ( \frac{kq}{r^2} ), ( \frac{2kq}{r^2} ). view available hint(s)

Explanation:

Part A

Step1: Analyze conductor's charge distribution

In a neutral conducting sphere with a cavity containing a charge \( +q \), the interior surface (cavity wall) must have an induced charge to cancel the electric field inside the conductor. By Gauss's law, the charge enclosed by a Gaussian surface inside the conductor (where \( E = 0 \)) must be zero. So the charge on the interior surface \( q_{\text{int}} \) and the central charge \( +q \) must sum to zero.

Step2: Solve for \( q_{\text{int}} \)

\( q + q_{\text{int}} = 0 \implies q_{\text{int}} = -q \)

Step1: Electric field in cavity (Gauss's law)

The cavity is a region with a point charge \( +q \) at its center. For a point charge, the electric field at distance \( r \) from it is given by Coulomb's law (or Gauss's law for a spherical symmetry). The formula for electric field due to a point charge \( q \) at distance \( r \) is \( E = \frac{kq}{r^2} \) (since \( k=\frac{1}{4\pi\epsilon_0} \)). The conducting material's charge doesn't affect the field inside the cavity (as the field inside conductor is zero, but cavity is a region with the point charge's field).

Step2: Determine the magnitude

The magnitude of the electric field \( E_{\text{int}} \) inside the cavity at distance \( r \) from the point charge is \( \frac{kq}{r^2} \) (since the field is radial and its magnitude follows Coulomb's law for the point charge, and the conductor's induced charges on the cavity wall and outer surface don't contribute to the field inside the cavity).

Answer:

\( -q \)

Part C