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

4

Explanation

Potential at a height H on the axis of the disc V(P). The charge dq contained on the ring shown in figure `dq = ( 2pi r dr ) sigma`
Potential at P due to this ring
`dv = (1)/(4 pi epsi_(0)) (dp)/(x)` where ` x = sqrt(H^(2) + x^(2))`
`therefore ` Potential due to the complete disc `V_(p) = int_(x = 0)^(x=2) dv = (sigma)/(2 epsi_(0)) int_( r =0)^( r =a)( r dr )/(sqrt(H^(2) + r^(2))`
`V_(p) = (sigma)/(2 epis_(0)) [ sqrt(a^(2) + H^(2)) - H]`
Potential centre (O) will be
(i) Particle is released from P and it just reaches point O. therefore, from conservation of mechanical energy Decrease in gravitational potential energy = Increase in electrostatic potential energy
`(Delta KE = 0 "because" k_(i) = k_(f) = 0) therefore m gH = q[V_(q) - V_(p)]`
` or gH = ((q)/(m)) ((simga)/(2 epis_(0))[a - sqrt(a^(2) + H^(2) + H] .....(i) `
Subsituting Eq. (i) We get
`gH = 2 g [ a + H - sqrt(a^(2) + H^(2)]` or `(H)/(2) = (a+H) - sqrt(a^(2) + H^(2))` or ` sqrt(a^(2) + H^(2))` =` a + (H)/(2)`
` or a^(2) + H^(2) = a^(2) + + (H^(2))/(4) = a or (3)/(4) H or H - = (4)/(3) a & H = 0 therefore H = ( 4//3)` Potential energy of the particle at height electrostatic potential energy + gravitational potential
energy `therefore " " U = qV + mqH :` Here V= Potenial at height H
`U = (sigma q)/( 2 epsi_(0)) [ sqrt(a^(2) + H^(2)) - H] + m gH " " ...(ii)`
At equilibrium position `F = (-dU)/(dH) = 0 , ` Differentitaing Eq . (ii) w.r.tH
`therefore mg + 2m g [(H)/( sqrt(a^(2) + H^(2))- 1] = 0 " " or " " 1 + (2H)/(sqrt(a^(2) + H^(2)) - 2 = 0`
`or (2H)/( sqrt(a^(2) + H^(2)) - 1 or (H^(2))/(a^(2) + H^(2)) = (1)/(4)`
or ` 3H^(2) - a^(2)`
`or H = (2)/(sqrt(3))`
From eq (ii) , we can write
U - E equations as
`U = mg ( 2 sqrt(a^(2) + H^(2)) - H)`
(Paabolic variation)
Therefore, graph will be as shown
Note that at `H - (a)/( sqrt(3))U` is minimum
Therefore `H = (a)/(sqrt(3))` U is is stable equilibrium position