Question

A thin disc of radius b=2a has a concentric hole of radius 'a' in it (see figure). It carries uniform surface charge $\sigma$ on it. If the electric field on its axis at height 'h' (h<<a) from its center is given as 'Ch' then value of 'C' is:

Answer 1

Uniform charged disk

$E=\frac{\sigma }{2E_0}[1-\frac{x}{(x^2+R^2{)}^{1/2}}]$

$=\frac{\sigma }{1E_0}[\sqrt{x^2+R^2}-x]$

Electric field due to complete disk $R=2a$ at distance $x$

$E_1=\frac{\sigma }{2E_0}[1-\frac{x}{(R^2+x^2{)}^{1/2}}(h=x;2a=R\left)\right]$

$E_1\frac{\sigma }{2E_0}[1-\frac{h}{(4a^2+h^2{)}^{1/2}}]=\frac{6}{2E_0}[1-\frac{h}{2a}]$

$\parallel$ by Electric that due to disk $R=a$

$E_2=\frac{\sigma }{2E_0}[1-\frac{h}{a}]$

now $E=E_1-E_2=\frac{\sigma }{2E_0}[1-\frac{h}{2a}]-\frac{\sigma }{2E_0}[1-\frac{h}{a}]=\frac{\sigma h}{4E_{0}a}$

$\therefore E=\frac{\sigma }{4\pi E_0}$