Q88 / 125 · BITSAT 2016 · Physics

The surface charge density of a thin charged disc of radius $\mathrm{R}$ is $\sigma$. The value of the electric field at the centre of the disc is $\frac{\sigma}{2 \epsilon_{0}}$. With respect to the field at the centre, the electric field along the axis at a distance $\mathrm{R}$ from the centre of the $\operatorname{disc}$

  • Correct answer
    A
    reduces by $70.7 \%$
  • B
    reduces by $29.3 \%$
  • C
    reduces by $9.7 \%$
  • D
    reduces by $14.6 \%$

Solution

Electric field intensity at the centre of the disc. \(\mathrm{E}=\frac{\sigma}{2 \epsilon_{0}} \text { (given) }\) Electric field along the axis at any distance \(x\) from the centre of the disc \(\mathrm{E}^{\prime}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}}-\mathrm{R}^{2}}\right)\) From question, \(\mathrm{x}=\mathrm{R}\) (radius of disc) \(\begin{aligned} &\therefore E^{\prime}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{R}{\sqrt{R^{2}+R^{2}}}\right) \\ &=\frac{\sigma}{2 \epsilon_{0}}\left(\frac{\sqrt{2} R-R}{\sqrt{2} R}\right) \\ &=\frac{4}{14} E \end{aligned}\) \(\therefore\) \% reduction in the value of electric field \(=\frac{\left(\mathrm{E}-\frac{4}{14} \mathrm{E}\right) \times 100}{\mathrm{E}}=\frac{1000}{14} \%=70.7 \%\)

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

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