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.

Explanation

(i) Given that a is the radius of disk, `sigma` is the surface charge density and `q//m = 4 epsilon_0 g// sigma`. The kinetic energy of the particle, when it reaches the disk, can be taken as zero. Potential due to a charged disk at any axial point situated at a distance x from O is
(##BMS_V03_C03_S01_047_S01##).
`V (x) = (sigma)/(2 epsilon_0) [sqrt(a^2 + x^2) - x]`
Hence,
`V (H) = (sigma)/(2 epsilon_0) [sqrt(a^2 + H^2) - H]`
and `V(O) = (sigma a)/(2 epsilon_0)`
According to the law of conservation of energy, we have the loss of graviatation potential energy is equal to the gain in electric potential energy.
`m g H = q Delta V = q [V(0) - V(H)]`
=`q[a - { sqrt((a^2 + H^2)) - H }] (sigma)/(2 epsilon_0)` ....(i)
we are given
`(sigma)/(2 epsilon_0) = 2 m g`
Putting this in Eq. (i), we get
`mg H = 2 mg [a - {sqrt((a^2 + H^2)) - H}]`
or `H = 2[a + H -sqrt((a^2 + H^2))]`
=`2 a + 2H - 2 sqrt((a^2 + H^2))`
or `2sqrt((a^2 + H^2)) = H + 2 a`
or `4a^2 + 4 H^2 = H^2 + 4 a^2 + 4 a H`
or `3 H^2 = 4 a H`
or `H = (4 a)/(3)` (Since H = 0 is not valid)
(ii) The total potential energy of the particle at height x is
`U (x) = mgx + qV (x) = mgx + (q sigma)/(2 epsilon _0) (sqrt(a^2 + x^2) - x )`
=`mgx + 2 mg + [sqrt((a^2 + x^2)) - x ]`
=`mg [2sqrt((a^2 + x^2)) - x]` ...(ii)
For equilibrium
`(d U)/(d x) = 0`
This gives `x = a// sqrt(3)` From Eq. (ii) graph between U(x) and x can be plotted as shown in (Fig. 3.99).
.