Q117 / 180 · AIIMS 2018 (26 May) · Physics

A uniformly charged non conducting disc with surface charge density $10 \mathrm{nC} / \mathrm{m}^2$ having radius $3 \mathrm{~cm}$. Then find the magnitude of electric field at a point on the perpendicular bisector at a distance $2 \mathrm{~cm}$ from origin of the disc. Figure

  • Correct answer
    A
    $251 \mathrm{~N} / \mathrm{C}$
  • B
    $126 \mathrm{~N} / \mathrm{C}$
  • C
    $502 \mathrm{~N} / \mathrm{C}$
  • D
    $72 \mathrm{~N} / \mathrm{C}$

Solution

$\begin{aligned} & \text { } E=\frac{\sigma}{4 \pi \varepsilon_0}(2 \pi)\left[1-\frac{x}{\sqrt{R^2+x^2}}\right] \\ & =9 \times 10^9 \times 10 \times 10^{-9} \times 6.28\left[1-\frac{2}{\sqrt{4+9}}\right] \\ & \therefore E=90 \times 6.28\left[1-\frac{2}{\sqrt{13}}\right]=251 \mathrm{~N} / \mathrm{C}\end{aligned}$

Hence, the correct answer is option: 'A'.

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