A thin disc of radius b = 2a has a concentric hole of radius 'a' in it (see figure). It carries uniform surface charge ' σ ' 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) (σ)/(aε₀) (b) (σ)/(2aε₀) (c) (σ)/(4aε₀) (d) (σ)/(8aε₀)

Answer: (c). Annulus = disc of radius 2a minus disc of radius a: E = (σ)/(2ε₀)[(h)/(√(h²+a²)) - (h)/(√(h²+4a²))]. For h a: E ≈ (σ h)/(2ε₀)(1/a - 1/2a) = (σ h)/(4aε₀) So C = (σ)/(4aε₀), answer (c).

A thin disc of radius b = 2a has a concentric hole of radius 'a' in… | JEE Main 2015 (10 Apr, Shift 1)

Q44JEE Main 2015 · 10 APR · SHIFT 1Difficult

Electric Field & Field Lines

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:

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Answer: (c)

Annulus = disc of radius $2a$ minus disc of radius $a$ : $E = \dfrac{\sigma}{2\varepsilon_0}\left[\dfrac{h}{\sqrt{h^2+a^2}} - \dfrac{h}{\sqrt{h^2+4a^2}}\right]$ .
For $h \ll a$ : $E \approx \dfrac{\sigma h}{2\varepsilon_0}\left(\dfrac{1}{a} - \dfrac{1}{2a}\right) = \dfrac{\sigma h}{4a\varepsilon_0}$
So $C = \dfrac{\sigma}{4a\varepsilon_0}$ , answer (c).

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Electric Field & Field Lines question from JEE Main 2015 (10 Apr, Shift 1)

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