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Standard XII

Physics

Electric Field Due to a Ring Along the Axis

A point charg...

Question

A point charge $Q$ is located on the axis of a disk of radius $R$ at a distance $b$ from the plane of the disk (Fig.). Show that if one-fourth of the electric flux from the charge passes through the disk, then $R = \sqrt{3} b$ .

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Solution

Solid angle subtended by disk at $Q$ ,
$Ω = 2 π (1 - cos α)$
$= 2 π [1 - \frac{b}{\sqrt{b^{2} + R^{2}}}]$
Flux through disk $= \frac{Q Ω}{ϵ_{0} 4 π}$
It is given to be $\frac{Q}{4 ϵ_{0}}$
So, $\frac{Q}{4 ϵ_{0}} = \frac{Q Ω}{ϵ_{0} 4 π}$
Solve to get $R = \sqrt{3} b$ .

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A point charge Q is located on the axis of a disk of radius R at a distance b from the plane of the disk (Fig.). Show that if one-fourth of the electric flux from the charge passes through the disk, then R=√3b.

Solid angle subtended by disk at Q,Ω=2π(1−cosα)=2π[1−b√b2+R2]Flux through disk =QΩϵ04πIt is given to be Q4ϵ0So, Q4ϵ0=QΩϵ04πSolve to get R=√3b.

A point charge $Q$ is located on the axis of a disk of radius $R$ at a distance $b$ from the plane of the disk (Fig.). Show that if one-fourth of the electric flux from the charge passes through the disk, then $R = \sqrt{3} b$ .

Solid angle subtended by disk at $Q$ ,
$Ω = 2 π (1 - cos α)$
$= 2 π [1 - \frac{b}{\sqrt{b^{2} + R^{2}}}]$
Flux through disk $= \frac{Q Ω}{ϵ_{0} 4 π}$
It is given to be $\frac{Q}{4 ϵ_{0}}$
So, $\frac{Q}{4 ϵ_{0}} = \frac{Q Ω}{ϵ_{0} 4 π}$
Solve to get $R = \sqrt{3} b$ .