A uniform surface charge density 8 $σ$ exists over the entire x-y plane except for a circular hole of radius a centered at the origin. The electric field at a point $P (0, 0, \sqrt{3} a)$ on the z- axis is found to be $\sqrt{x} \frac{σ}{ε_{0}}$ . Find the value of x.

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answer is 12.

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Detailed Solution

field due to entire sheet: $E_{T} = \frac{8 σ}{2 ε_{0}}$

field due to sheet with the hole:

$E_{net} = E_{T} - E_{disc}$

$\Rightarrow E_{net} = \frac{8 σ}{2 ε_{0}} - \frac{8 σ}{2 ε_{0}} (1 - \frac{z}{\sqrt{z^{2} + a^{2}}})$

$\Rightarrow E_{net} = \frac{8 σz}{2 ε_{0} \sqrt{z^{2} + a^{2}}}$

$\Rightarrow E_{net} = \frac{σ}{ε_{0}} \frac{4 \sqrt{3} a}{2 a} = \frac{σ}{ε_{0}} \sqrt{12}$

$\Rightarrow x = 12$

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