A non-conducting disc of radius a with 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 disc from a height H with zero initial velocity. Charge per unit mass of the particle is `q/m = (4 epsilon_0 g)/(sigma)`.
(ii) Find the height for its equilibrium position.

Explanation

Let us assume that the amount of charge in the ring of thickness dr is dq.
Potential at P due to this ring,
`dV = 1/(4 pi epsilon_0) (dq)/(x)`, where `x = sqrt(H^(2)+r^(2))`
`:. dV = 1/(4 pi epsilon_0) ((2pi r dr)sigma)/(sqrt(H^(2)+r^(2))) [ :' dq = (2 pi r dr)sigma]`
`=(sigma)/(2epsilon_0) (rdr)/(sqrt(H^(2)+r^(2)))`

Now, potential at P due to the complete disc,
`V_(p) = int_(r=0)^(r=a)dV=(sigma)/(2epsilon_0)int_(r=0)^(r=a) (rdr)/(sqrt(H^(2)+r^(2)))`
`:. V_(P) = (sigma)/(2epsilon_0)[sqrt(a^(2)+H^(2))-H]`
So potential at the centre `(O) , V_(O) = (sigma a)/(2 epsilon_0) [ :' H = 0 ]`
(ii) Potential energy of the particle at height H = electrostatic potential energy + gravitational potential energy
or, `U = qV_(P) + mgH [ V_(P` = electric potential at high H]
or, `U = (sigma q)//(2epsilon_0) [sqrt(a^(2)+H^(2))-H] + mgH` ...........(2)
At equilibrium , `F = -(dU)/(dH) = 0`
DIfferntiating equation, (2) with respect to H,
`mg + sigma q/(2 epsilon_0)[(1/2)(2H)1/(sqrt(a^(2)+H^(2)))-1] = 0`
or, `mg + 2mg [(H)/(sqrt(a^(2)+H^(2)))-1] = 0 [ :' (sigma q)/(2 epsilon_0) = 2mg]`
or, `1 + (2H)/(sqrt(a^(2)+H^(2))) -2 = 0 or, 4H^(2)-H^(2)=a^(2)`
`:. H = a/(sqrt3)`.