JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A thin disc of radius \(b = 2a\) has a concentric hole of radius \('a'\) in it (see figure). It carries uniform surface charge \('\sigma '\) on it. If the electric field on its axis at height \('h'\) \((h << a)\) from its centre is given as \('Ch'\) then value of \('C'\) is
- A \(\frac{\sigma }{{4a{ \in _0}}}\)
- B \(\frac{\sigma }{{8a{ \in _0}}}\)
- C \(\frac{\sigma }{{a{ \in _0}}}\)
- D \(\frac{\sigma }{{2a{ \in _0}}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sigma }{{4a{ \in _0}}}\)
Step-by-step Solution
Detailed explanation
Eleatric field due to complete disc \((R=2 a)\) at a distance \(x\) and on its axis \(E_{1}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{x}{\sqrt{R^{2}+x^{2}}}\right]\) \(E_{1}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{\sqrt{4 a^{2}+h^{2}}}\right]\)…
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