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(A) The electric field $E$ on the axis of a disc of radius $R$ at a distance $h$ from its center is given by $E = \frac{\sigma}{2\epsilon_0} \left( 1

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$\frac{\sigma}{4a\epsilon_0}$

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(A) The electric field $E$ on the axis of a disc of radius $R$ at a distance $h$ from its center is given by $E = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{\sqrt{R^2 + h^2}} \right)$.For the disc with a hole,we can consider it as a large disc of radius $R_1 = 2a$ minus a smaller disc of radius $R_2 = a$.The electric field $E_1$ due to the large disc is $E_1 = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{\sqrt{(2a)^2 + h^2}} \right) = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{2a \sqrt{1 + (h/2a)^2}} \right)$.Since $h The electric field $E_2$ due to the smaller disc is $E_2 = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{\sqrt{a^2 + h^2}} \right) \approx \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{a} \right)$.The net electric field $E = E_1 - E_2 = \frac{\sigma}{2\epsilon_0} \left[ (1 - \frac{h}{2a}) - (1 - \frac{h}{a}) \right] = \frac{\sigma}{2\epsilon_0} \left( \frac{h}{a} - \frac{h}{2a} \right) = \frac{\sigma}{2\epsilon_0} \left( \frac{h}{2a} \right) = \frac{\sigma h}{4a\epsilon_0}$.Comparing this with $Ch$,we get $C = \frac{\sigma}{4a\epsilon_0}$.

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

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