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JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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
- A reduces by \(70. 7\%\)
- B reduces by \(29.3\%\)
- C reduces by \(9.7\%\)
- D reduces by \(14.6\%\)
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 \(E^{\prime}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{x}{\sqrt{x^{2}+R^{2}}}\right)\)…
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