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Q.

A very thin disc is uniformly charged with surface charge density $σ > 0$ . Then the electric field intensity on the axis at the point from which the disc is seen at an solid angle $Ω$ is

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a

3. $\frac{1}{4 {πε}_{0}} Ω$

b

1. $\frac{1}{2 {πε}_{0}} σ Ω$

c

4.None

d

2. $\frac{1}{4 {πε}_{0}} σ Ω$

answer is B.

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Detailed Solution

Net $\vec{E}$ field will be along axis $C D$

So,

$d E_{x} = d E \cos θ$

$= \frac{k d q}{r^{2}} \cos θ$

$= k \frac{σ d S}{r^{2}} \cos θ (d q = σ d s)$

$= κ σ \frac{d s \cos θ}{r^{2}}$

$= κ σ d Ω$

$(d Ω = s o l i d a n g l e = \frac{d s \cos θ}{r^{2}})$

$\vec{E} = \int d E_{x} = \int_{0}^{Ω} k σ d Ω$

$E_{x} = κ σ Ω$

$E_{x} = \frac{1}{4 {πε}_{0}} σ Ω$

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