91Ó°ÊÓ

• The charge of the particle,$q=1.0×{10}^{-7}C.$
• The spherical cavity of radius, r = 3.0 cm.

To explain the electrostatic force between the two charges, we assume that the charges create an electric field around them. The magnitude of electric field E set up by the electric charge q at a distance r is given as,

$E=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{\mathrm{r}}^2}$

Using the concept of the electric field to determine the magnitude of the electric field

The directionof the electric field at$p_1$is away from$q_1$, and its magnitude is

$$\begin{gathered} E=\frac{q}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2} \\ =\frac{(9×{10}^{9}N.m^2/c^2)(1.0×{10}^{-7}C)}{{(0.015m)}^2} \\ =4.0×{10}^{6}N/C \end{gathered}$$

The electric field at 1.5 cm from the cavity center is$4.0×{10}^{6}N/C.$

From Gauss law, the charge enclosed inside the metal is zero. The point$P_2$is inside the metal. Therefore, electric fieldE, due to the electric charge inside the metal, is zero.