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

Answer 1

The correct option is A $\frac{\sigma y}{2{ϵ}_0}[\frac{1}{\sqrt{R_1^2+y^2}}-\frac{1}{\sqrt{R_2^2+y^2}}]$


Consider a hypothetical ring of radius $x$ and thickness $dx$ on the disc.
The charge on the hypothetical ring :
$dq=\sigma .2\pi xdx$

Now the electric field at point P due to the ring is :

$dE=\frac{dq}{4\pi {ϵ}_0}\times \frac{y}{(x^2+y^2{)}^{3/2}}$​

$dE=\frac{\sigma .2\pi xdx}{4\pi {ϵ}_0}\times \frac{y}{(x^2+y^2{)}^{3/2}}$​

For complete annular disc,
$E=\int dE=\frac{\sigma y}{2{ϵ}_0}\int \frac{x}{(x^2+y^2{)}^{3/2}}dx$​

Let,
$x^2+y^2=p^2\Rightarrow 2xdx=2pdp$

$\Rightarrow \int \frac{x}{(x^2+y^2{)}^{3/2}}dx=\int \frac{pdp}{p^3}=-\frac{1}{p}$​

$\therefore E=\frac{\sigma y}{2{ϵ}_0}{\left[\frac{-1}{\sqrt{x^2+y^2}}\right]}_{R_1}^{R_2}$

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