Question

A particle of charge $\mathbf{q}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{C}$is at the center of a spherical cavity of radius $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{cm}$in a chunk of metal. Find the electric field

(a)$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{cm}$from the cavity center and

(b) anyplace in the metal.

Short answer

(a) The electric field at 1.5 cm from the cavity center is$4.0\times {10}^6\mathrm{N}/\mathrm{C}.$

(b) The electric field at anyplace in the metal is zero.

Step-by-step solution

Listing the given quantities

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

Understanding the concept of Electric field

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,

$E=\frac{q}{4\pi {\epsilon }_0{\mathrm{r}}^2}$

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

(a) Calculations for the electric field at 1.5 cm from the cavity center

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

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

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

(b) Calculations for the electric field at any place inside the metal

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.