# A thin disc of radius b=2a has a concentric hole of radius 'a' in it (see figure). It carries uniform surface charge 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 : Option 1) Option 2) Option 3) Option 4)

As we discussed in

Uniformly charged disc -

$\dpi{100} E=\frac{\sigma }{2\epsilon _{0}}\left [ 1-\frac{x}{\left ( x^{2}+R^{2} \right )^{\frac{1}{2}}} \right ]$

$\dpi{100} V=\frac{\sigma }{2\epsilon _{0}}\left [ \sqrt{x^{2}+R^{2}} -x\right ]$

- wherein

Electric Field due to complete disc (R = 2a) at distance x

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

$E_1=\frac{\sigma}{2 \epsilon_o}\left [ 1 - \frac{h}{(4{a}^2+h^2)^\frac{1}{2}} \right ] = \frac{\sigma}{2 \epsilon_o}\left [ 1 - \frac{h}{2a} \right ]$

Similarly electric field due to disc (R = a )

$E_2=\frac{\sigma}{2 \epsilon_o}\left [ 1 - \frac{h}{a} \right ]$

$Now \: E = E_1 - E_2 = \frac{\sigma}{2 \epsilon_o}\left [ 1 - \frac{h}{2a}\right ]-\frac{\sigma}{2 \epsilon_o}\left [ 1 - \frac{h}{a}\right ] = \frac{\sigma h }{4\epsilon_o a}$

$Hence \: C = \frac{\sigma}{4 \epsilon_o a}$

Option 1)

Incorrect

Option 2)

Incorrect

Option 3)

Correct

Option 4)

Incorrect

N

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