A disc having uniform surface charge density σ is placed in y-z with center at origin and radius R.
Find the electric field at a point (x, 0, 0).
(a) Ex = \(\frac{σ}{2\in_0}\)\(\Big[ 1 -\frac{x}{\sqrt{R^2 + x^2}}\Big]\)
(b) Ex = \(\frac{σ}{\in_0}\)\(\Big[ 1 -\frac{\sqrt{R^2 + x^2}}{x}\Big]\)
(c) Ex = \(\frac{σ}{\in_0}\)\(\Big[ 1 -\frac{x}{\sqrt{R^2 + x^2}}\Big]\)
(d) Ex = \(\frac{σ}{2\in_0}\)\(\Big[x -\frac{x}{\sqrt{R^2 + x^2}}\Big]\)
Correct answer: (a) Ex = \(\frac{σ}{2\in_0}\)\(\Big[ 1 -\frac{x}{\sqrt{R^2 + x^2}}\Big]\)
Explanation:
The disc can be considered to be a collection of a large number of concentric rings. Consider an element of the shape of rings of radius r and of width dr. Electric field due to this ring at P is;
dE = \(\frac{K.σ^2 π r.dr.x}{(r^2+x^2)^{\frac{3}{2}}}\)
Put, r2 + x2 = y2
2rdr = 2ydy
∴ dE = \(\frac{K.σ^2 \pi y.dy.x}{y^3}\)
= 2Kσx.x\(\frac{ydy}{y^3}\)
∴ E = ∫dE ⟹ E = 2Kσπx \(\int_x^x \frac{1}{y^2}\)dy
= 2Kσπx.\(\Big[-\frac{1}{y}\Big]_x\)
= 2Kσπx\(\Big[\frac{1}{x} -\frac{x}{\sqrt{R^2 + x^2}} \Big]\)
= 2Kσx \(\Big[ 1 -\frac{x}{\sqrt{R^2 + x^2}}\Big]\)
= \(\frac{σ}{2\in_0}\)\(\Big[ 1 -\frac{x}{\sqrt{R^2 + x^2}}\Big]\) along the axis.