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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2013
  • Hard
  • MCQ

The surface charge density of a thin charged disc of radius \(R\) is \(\sigma \). The value of the electric field at the centre of the disc is \(\frac{\sigma }{{2\,{ \in _0}}}\). With respect to the field at the centre, the electric field along the axis at a distance \(R\) from the centre of the disc

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Answer & Solution

Correct Answer

(A) reduces by \(70. 7\%\)

Step-by-step Solution

Detailed explanation

Electric field intensity at the centre of the disc.

\(E=\frac{\sigma}{2 \epsilon_{0}} \quad(\text { given })\)

Electric field along the axis at any distance

\(x\) from the centre of the disc

\(E^{\prime}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{x}{\sqrt{x^{2}+R^{2}}}\right)\)

From question, \(x=R(\text { radius of disc })\)

\(\therefore \mathrm{E}^{\prime}=\) \( \frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{R}^{2}}}\right) \)

\(= \frac{\sigma}{2 \epsilon_{0}}\left(\frac{\sqrt{2} \mathrm{R}-\mathrm{R}}{\sqrt{2} \mathrm{R}}\right) \)

\(=\frac{4}{14} \mathrm{E}\)

\(\therefore\) \(\%\) reduction in the value of electric field

\(=\frac{\left(\mathrm{E}-\frac{4}{14} \mathrm{E}\right) \times 100}{\mathrm{E}}=\frac{1000}{14} \%=70.7 \%\)

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