Question 20.5: An infinite sheet of charge is lying on the xy-plane as show......
An infinite sheet of charge is lying on the xy-plane as shown in Fig. 20.16. A positive charge is distributed uniformly over the plane of the sheet with a charge per unit area σ. Calculate the electric field at a point P located a distance a from the plane.
Let us divide the plane into narrow strips parallel to the y-axis.
A strip of width dx can be considered as an infinitely long wire of charge per unit length λ = σ dx. From Eq. 20.36,
E = \underset{L \rightarrow \infty }{lim} \frac{2 k \lambda}{a \sqrt{(2 a /L)^2+1}} \Rightarrow E = 2k \frac{\lambda}{a} (20.36)
at point P, the strip sets up an electric field d\vec{E} lying in the xz-plane of magnitude:
d E=2k{\frac{\lambda}{r}}=2k{\frac{\sigma d x}{r}}This electric field vector can be resolved into two components d\vec{E}_{x} and d\vec{E}_{z}.
By symmetry the components d\vec{E}_{x} will sum to zero when we consider the entire sheet of charge. Therefore, the resultant electric field at point P will be in the z-direction, perpendicular to the sheet. From Fig. 20.16, we find the following:
and hence:
E=\int d E_{z}=2k{\sigma}\int\limits_{-\infty}^{+\infty}\frac{\sin\theta d x}{r}To perform the integration of this expression, we must first relate the variables θ , x, and r. One approach is to express θ and r in terms of x. From the geometry of Fig. 20.16, we have:
r={\sqrt{x^{2}+a^{2}}}\quad{\mathrm{and}}\quad\sin\theta={\frac{a}{r}}={\frac{a}{\sqrt{x^{2}+a^{2}}}}Then, from the table of integrals in Appendix B, we find that:
E=2k\sigma a\int\limits_{-\infty}^{+\infty}{\frac{d x}{x^{2}+a^{2}}}=2k\sigma a\left|{\frac{1}{a}}\tan^{-1}{\frac{x}{a}}\right|_{-\infty}^{+\infty}=2k\sigma\left[\tan^{-1}(\infty)-\tan^{-1}(-\infty)\right]=2k\sigma~\left[{\frac{\pi}{2}}+{\frac{\pi}{2}}\right]
Thus:
E=2\pi k\sigma={\frac{\sigma}{2\epsilon_{o}}}This result is identical to the one we shall find in Sect. 20.4.4 for a charged disk of infinite radius. We note that the distance a from the plane to the point P does not appear in the final result of E. This means that the electric field set up at any point by an infinite plane sheet of charge is independent of how far the point is from the plane. In other words, the electric field is uniform and normal to the plane.
Also, the same result is obtained if the point P lies below the xy-plane. That is, the field below the plan has the same magnitude as that above the plane but as a vector it points in the opposite direction.