A nonconducting disk of radius a and uniform positive surface charge density `sigma` is placed on the ground, with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disk, from a height H with zero initial velocity. The particle has `q//m = 4 epsilon_(0) g// sigma`.
(i) Find the value of H if the particle just reaches the disk.
(ii) Sketch the potential energy of the particle as a function of its height and find its equilibrium position.

Answer

A, B, C, D

Explanation

(a) KEY CONCEPT: The K.E. of the particle, when it reaches the disc is zero.
Given that a=radius of disc, `sigma=` surface charge density, `q//m=4epsilon_0g//sigma`
Potential due to a charged disc at any axial point situated at a distance x from O is,
`V(x)=(sigma)/(2epsilon_0)[sqrt(a^2+x^2)-x]`
Hence, `V(H)=(sigma)/(2epsilon_0)[sqrt(a^2+H^2)-H]`
and `V(O)=(sigmaa)/(2epsilon_0)`

`mgH=qDeltaV`
`=q[V(0)-V(H)]`
`mgH=q(sigma)/(2epsilon_0)[a-{sqrt(a^2+H^2)-H}]` ...(1)
From the given relation: `(sigmaq)/(2epsilon_0)=2mg`
Putting this in eqution (1), we get,
`mgH=2mg[a-{sqrt(a^2+H^2)-H}]`
or `H=(4a)/(3)` [ `:'` H=O is not valid]
(b) Total potential energy of the particle at height H `U(x)=mgx+qV(x)`
`=mgx+(qsigma)/(2epsilon_0)(sqrt(a^2+x^2)-x)`
`=mgx+2mg[sqrt((a^2+x^2))-x]` ...(2)
`U_((A))=mgH+2mg[sqrt(a^2+H^2)-H]`
`=mg[2sqrt(a^2+H^2)-H^2]` ...(3)
For equilibrium : `(dU)/(dH)=0`
This gives: `H=a/sqrt3 :. U_(min)=sqrt3 mga`
From equation (2), graph between U(x) and x is as shown above.