Question 127 of 311

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 a height $\text{h}\left(\text{h}\lt \lt \text{a}\right)$ from its centre is given as $\text{C}\text{h}$ then the value of $C$ is

A$\dfrac{\sigma }{4a{ϵ}_{0}}$B$\dfrac{\sigma }{a{ϵ}_{0}}$C$\dfrac{\sigma }{5a{ϵ}_{0}}$D$\dfrac{\sigma }{2a{ϵ}_{0}}$

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Answer: (A) $\dfrac{\sigma }{4a{ϵ}_{0}}$

$∵$ At the axial point of a uniformly charged disc electric field is given by,

$E=\dfrac{\sigma }{2{ϵ}_{0}}\left(1-cos\theta \right)$



By superposition principle, when inner disc is removed , then, electric field due to remaining disc is,



$E=\dfrac{\sigma }{2{ϵ}_{0}}\left[\left(1-cos{\theta }_{2}\right)-\left(1-cos{\theta }_{1}\right)\right]$

$=\dfrac{\sigma }{2{ϵ}_{0}}\left[cos{\theta }_{1}-cos{\theta }_{2}\right]$

$=\dfrac{\sigma }{2{ϵ}_{0}}\left[\dfrac{h}{\sqrt{{h}^{2}+{a}^{2}}}-\dfrac{h}{\sqrt{{h}^{2}+{b}^{2}}}\right]$

$=\dfrac{\sigma }{2{ϵ}_{0}}\left[\dfrac{h}{a\sqrt{1+\dfrac{{h}^{2}}{{a}^{2}}}}-\dfrac{h}{b\sqrt{1+\dfrac{{h}^{2}}{{b}^{2}}}}\right]$

$∵h≪a$ and b.

$∴E=\dfrac{\sigma }{2{ϵ}_{0}}\left[\dfrac{h}{a}-\dfrac{h}{b}\right]$

$=\dfrac{\sigma }{2{ϵ}_{0}}\left[\dfrac{h}{a}-\dfrac{h}{2a}\right]=\dfrac{\sigma h}{4{ϵ}_{0}a}$

$\Rightarrow C=\dfrac{\sigma }{4{aϵ}_{0}}$.