Question

A thin disc of radius $b=2a$ has 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 centre is given as ‘$Ch$’, then value of ‘$C$’ is:

• A. $\frac{\sigma }{2a{\epsilon }_0}$
• B. $\frac{\sigma }{4a{\epsilon }_0}$
• C. $\frac{\sigma }{8a{\epsilon }_0}$
• D. $\frac{\sigma }{a{\epsilon }_0}$

Answer 1

The correct option is B $\frac{\sigma }{4a{\epsilon }_0}$

Electric field at an axial point of a charged disc at a distance $x$ is
$E=\frac{\sigma }{2{\epsilon }_0}[1-\frac{x}{\sqrt{R^2+x^2}}]$

Electric field due to the complete disc of $R=2a$ at height $h$,
$E_1=\frac{\sigma }{2{\epsilon }_0}[1-\frac{h}{\sqrt{4a^2+h^2}}]$

$=\frac{\sigma }{2{\epsilon }_0}[1-\frac{h}{2a}]$ For $[h<<a]$

Electric field due to the disc of $R=a$ at height $h$,
$E_2=\frac{\sigma }{2{\epsilon }_0}[1-\frac{h}{a}]$

Electric field due to the given annular disc,
$E=E_1-E_2=\frac{\sigma h}{4{\epsilon }_{0}a}$