Question

An annular disc has inner and outer radius R${}_1$ and R${}_2$ respectively. Charge is uniformly distributed. Surface charge density is $\sigma$. Find the electric field at any point distant y along the axis of the disc.

• A. $\frac{\sigma }{2{\epsilon }_0}$
• B. $\frac{\sigma y}{2{\epsilon }_0(R_2-R_1)}$
• C. $\frac{\sigma y}{2{\epsilon }_0}[\frac{1}{\sqrt{R_1^2+y^2}}-\frac{1}{\sqrt{R_2^2+y^2}}]$
• D. $\frac{\sigma }{2{\epsilon }_0}log\frac{R_2+y}{R_1+y}$

Answer 1

Answer: (C)

$\frac{\sigma y}{2{\epsilon }_0}[\frac{1}{\sqrt{R_1^2+y^2}}-\frac{1}{\sqrt{R_2^2+y^2}}]$

Consider a hypothetical ring of radius $x$ and thickness $dx$. The charge on the hypothetical ring, $dq=\sigma .2\pi x$ . Now the electric field at point P due to the ring is
$dE=\frac{dq}{4\pi {ϵ}_0}.\frac{y}{(x^2+y^2{)}^{3/2}}=\frac{\sigma .2\pi x}{4\pi {ϵ}_0}.\frac{y}{(x^2+y^2{)}^{3/2}}$
For disc,
$E=\frac{\sigma y}{2{ϵ}_0}{\int }_{R_1}^{R_2}\frac{x}{(x^2+y^2{)}^{3/2}}dx$
let $x^2+y^2=p^2,2xdx=2pdp,\int \frac{x}{(x^2+y^2{)}^{3/2}}dx=\int \frac{pdp}{p^3}=-\frac{1}{p}=-\frac{1}{\sqrt{x^2+y^2}}$
now $E=\frac{\sigma y}{2{ϵ}_0}{[-\frac{1}{\sqrt{x^2+y^2}}]}_{R_1}^{R_2}=\frac{\sigma y}{2{ϵ}_0}[-\frac{1}{\sqrt{R_2^2+y^2}}+\frac{1}{\sqrt{R_1^2+y^2}}]$