Question

In Fig. 23-45, a small circular hole of radius$\mathit{R}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{}\mathit{c}\mathit{m}$has been cut in the middle of an infinite, flat, non-conducting surface that has uniform charge density$\mathbf{\sigma }\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{pC}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. A z-axis, with its origin at the hole’s center, is perpendicular to the surface. In unit-vector notation, what is the electric field at point Pat$\mathbf{z}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{56}\mathbf{}\mathbf{cm}$? (Hint:See Eq. 22-26 and use superposition.)

Short answer

The electric field at point P at z=2.56 cm is$\left(0.208\mathrm{N}/\mathrm{C}\right)\stackrel{⏜}{\mathrm{k}}$ .

Step-by-step solution

The given data

a) The radius of the small circular hole, R =1.80 cm

b) Surface charge density,$\mathrm{\sigma }=4.50\mathrm{pC}/{\mathrm{m}}^2$

c) Distance of the point,z =2.56 cm

Understanding the concept of the electric field

Using the concept of Gauss law-planar symmetry of the electric field of a non-conducting sheet, we can get the required value of the net electric field by substituting the electric field for sheet and pad as calculated.

Formula:

The electric field of a non-conducting sheet, $\mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}$ (1)

Calculation of the electric field

The charge distribution in this problem is equivalent to that of an infinite sheet of charge with surface charge density $\mathrm{\sigma }=4.50\mathrm{pC}/{\mathrm{m}}^2$. Again, a small circular pad of radiusR=1.80 cm located at the middle of the sheet with charge density$-\sigma$. The electric fields produced by the sheet and the pad with subscripts 1 and 2, respectively. Using for$E_2$, the net electric field $E$at a distance z=2.56 cm along the central axis is then given using equation (1) as given:

$$\begin{gathered} \overrightarrow{\mathrm{E}}={\overrightarrow{\mathrm{E}}}_1+{\overrightarrow{\mathrm{E}}}_2 \\ =\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left(\stackrel{⏜}{\mathrm{k}}\right)+\frac{\left(-\mathrm{\sigma }\right)}{2{\mathrm{\epsilon }}_0}\left(1-\frac{\mathrm{z}}{\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\right)\stackrel{⏜}{\mathrm{k}} \\ =\frac{\mathrm{\sigma z}}{2{\mathrm{\epsilon }}_0\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\stackrel{⏜}{\mathrm{k}} \\ =\frac{\left(4.50\times {10}^{-12}\mathrm{C}/{\mathrm{m}}^2\right)\left(2.56\times {10}^{-2}\mathrm{m}\right)}{2\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\sqrt{{\left(2.56\times {10}^{-2}\mathrm{m}\right)}^2+{\left(1.80\times {10}^{-2}\mathrm{m}\right)}^2}} \\ =\left(0.208\mathrm{N}/\mathrm{C}\right)\stackrel{⏜}{\mathrm{k}} \end{gathered}$$

Hence, the value of the electric field is $\left(0.208\mathrm{N}/\mathrm{C}\right)\stackrel{⏜}{\mathrm{k}}$.