91Ó°ÊÓ

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

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

c) Distance of the point,z =2.56 cm

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, $\mathrm{E}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}$ (1)

The charge distribution in this problem is equivalent to that of an infinite sheet of charge with surface charge density $\mathrm{σ}=4.50\mathrm{pC}/{\mathrm{m}}^2$. Again, a small circular pad of radiusR=1.80 cm located at the middle of the sheet with charge density$-\mathrm{σ}$. 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:

$$\begin{gathered} \stackrel{→}{\mathrm{E}}={\stackrel{→}{\mathrm{E}}}_1+{\stackrel{→}{\mathrm{E}}}_2 \\ =\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\left(\stackrel{Áåœ}{\mathrm{k}}\right)+\frac{\left(-\mathrm{σ}\right)}{2{\mathrm{Î\mu }}_0}\left(1-\frac{\mathrm{z}}{\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\right)\stackrel{Áåœ}{\mathrm{k}} \\ =\frac{\mathrm{σ³ú}}{2{\mathrm{Î\mu }}_0\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\stackrel{Áåœ}{\mathrm{k}} \\ =\frac{\left(4.50×{10}^{-12}\mathrm{C}/{\mathrm{m}}^2\right)\left(2.56×{10}^{-2}\mathrm{m}\right)}{2\left(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\sqrt{{\left(2.56×{10}^{-2}\mathrm{m}\right)}^2+{\left(1.80×{10}^{-2}\mathrm{m}\right)}^2}} \\ =\left(0.208\mathrm{N}/\mathrm{C}\right)\stackrel{Áåœ}{\mathrm{k}} \end{gathered}$$

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