(i) H = 4 a/3

(ii) U(y) = 2mg + mgy; at equilibrium y =

Sol. Potential at a height H on the axis of the disc ie. V(P): →

The charge dq contained in the ring shown in figure

dq = (2πrdr)σ

Potential at P due to this ring,

∴ Potential due to the complete disc,

=

or, V p =

Potential at 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

( Δ KE = 0 because K i = K f = 0)

∴ mgH = q [ V o – V p ]

Substituting in (1), we get

gH = 2g [a + H –

or = (a + H) –

∴ H = (4/3)a

(ii) Potential energy of the particle at height H = Electrostatic potential energy + gravitational potential energy

∴ U = qV + mgH

Here V = Potential at height H

U–H equation as

U = mg

∴ U = 2mga at H = 0 and

U = U min = mga at H =

Therefore U–H graph will be as shown.

Note that at H = , U is minimum.

Therefore, H = is stable equilibrium position.