Question 23.34: In Fig. 23-45, a small circular hole of radius R = 1.80 cm h......
In Fig. 23-45, a small circular hole of radius R = 1.80 cm has been cut in the middle of an infinite, flat, nonconducting surface that has uniform charge density σ = 4.50 pC/m². 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 P at z = 2.56 cm? (Hint: See Eq. 22-26 and use superposition.)
E=\frac{\sigma}{2 \varepsilon_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right), (22-26)
The charge distribution in this problem is equivalent to that of an infinite sheet of charge with surface charge density \sigma=4.50 \times 10^{-12}\, C / m ^2 plus a small circular pad of radius R = 1.80 cm located at the middle of the sheet with charge density –σ. We denote the electric fields produced by the sheet and the pad with subscripts 1 and 2, respectively. Using Eq. 22-26 for \vec{E}_2 ,the net electric field \vec{E} at a distance z = 2.56 cm along the central axis is then
\begin{aligned} \vec{E} & =\vec{E}_1+\vec{E}_2=\left(\frac{\sigma}{2 \varepsilon_0}\right) \hat{ k }+\frac{(-\sigma)}{2 \varepsilon_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right) \hat{ k }=\frac{\sigma z}{2 \varepsilon_0 \sqrt{z^2+R^2}} \hat{ k } \\ & =\frac{\left(4.50 \times 10^{-12} \,C / m ^2\right)\left(2.56 \times 10^{-2} \,m \right)}{2\left(8.85 \times 10^{-12} \,C ^2 / N \cdot m ^2\right) \sqrt{\left(2.56 \times 10^{-2}\, m \right)^2+\left(1.80 \times 10^{-2} \,m \right)^2}} \hat{ k }=(0.208 \,N / C ) \hat{ k }. \end{aligned}