Question

quick quiz 22.1
suppose a point - charge is located at the center of a spherical surface. the electric field at the surface of the sphere and the total flux through the sphere are determined. now the radius of the sphere is halved. what happens to the flux through the sphere and the magnitude of the electric field at the surface of the sphere?
(a) the flux and field both increase.
(b) the flux and field both decrease.
(c) the flux increases, and the field decreases.
(d) the flux remains the same, and the field increases.
(e) the flux decreases, and the field increases.
(f) the flux decreases, and the field remains the same.

Explanation:

Step1: Recall Gauss's law

$\varPhi_E=\frac{q_{enclosed}}{\epsilon_0}$, where $\varPhi_E$ is electric - flux, $q_{enclosed}$ is the charge enclosed and $\epsilon_0$ is the permittivity of free space. Since the charge at the center is constant, the flux $\varPhi_E$ remains the same as it depends only on the enclosed charge.

Step2: Recall electric - field formula

$E = \frac{kq}{r^2}$, where $k$ is Coulomb's constant, $q$ is the source charge and $r$ is the distance from the source charge. As the radius $r$ is halved, the electric - field $E$ increases.

Answer:

(d) The flux remains the same, and the field increases.