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A point charge Q is located on the axis ...

A point charge `Q` is located on the axis of disc of a radius `R` at a distance a from the plane of the disc . If one fourth `(1//4th)` of the flus from the charge passes through the disc, then find the relation between a `& R `.
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Text Solution

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The correct Answer is:
` a= R/(sqrt 3)`
Flux through ring ` d phi = E dA cos theta `
` d phi _E = 1/( 4 pi in_0) Q/((a^2 = x^2)) . 2 pi x d x . a/( sqrt a^2 + x^2)`

`phi _E = 1/(4 in_0) Qa int_0^R (2 x d x)/(( a^2 + x^2 )^(3//2))` ….(1)
Put `a^2 + x^2 =t`
From (1) and (2)
`phi_E = 1/(4 phi_0) Qa int_0^R (dt)/(t^(3//2)) = (Qa)/(4in_0) [-2/(sqrt t) ]_0^r = (Qa)/(4 in_0) [- 2/(sqrt a^@ + R^2)]_0^R`.
`phi_E=(QWa)/(2in_0) [1/a - 1/(sqrt ( a^2 +r^2)] =n Q/(2 in_0) (1- a/(sqrt a^2 + R^2))`
`phi _E = Q/(2 in_0) (1- a/(sqrt (R^2 +a^2))) = (phi_("toal"))/2 (1 - a/(sqrt (R^2 +a^2)))`
`1/4 = 1/2 (1-n a/(sqrt R^2 + a^2 ))`
`1/2 = 1 - a/( sqrt (R^2 +a^2)) rArr a/( sqrt (R^2 +a^2 )) = 1/2`
`4 a^2 = R^2 +a^2 rArr a= R/(sqrt 3` .
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