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

Potential at a height H on the axis of the disc `V(P)`. The charge `dq` contained in the ring shown in figure `dq=(2pirdr)sigma`
Potential of P due to this ring
`dV=1/(4epsilon_0).(dq)/x`, where `x=sqrt(H^2-r^2)`
`dV=1/(4piepsilon_0).((2pirdr)sigma)/sqrt(H^2+r^2)=sigma/(2epsilon_0)(rdr)/sqrt(H^2+r^2)`
`:.` Potential 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]`
Potential at centre, O will be
`V_O=(sigmaa)/(2epsilon_0)` `(H=0)`
a. Particle is relesed from P and it just reaches point O. Therefore, from conservation of mechanical energy
decrease is gravtational potential energy = increase in electrostatic potential energy
`(/_\KE=0` because `K_i=K_f=0)`
`:.mgH=q[V_O-V_P]`
or `gH=(q/m)(sigma/(2epsilon_0)[a-sqrt(a^2+H^2)+H]`
`q/m=(4epsilon_0g)/sigmaimplies(qsigma)/(2epsilonm)=2g`
Substituting in Eqn. i we get
`gH=2g[a+H-sqrt(a^2+H^2)]`
or `H/2 =(a+H)-sqrt(a^2+H^2)`
or `sqrt a^2+H^2=a^2+H^2/4+aH`
or `3/4=aH` or `H=4/3 a ` and `H=0`
`:. =(4//3)a`
b. Potential energy of the particle at height `H=` Electrostatic potential energy + gravitational potentail energy
`:. U=qV+mgH`
Here, `V=` Potential at height H
`U=(sigmaq)/(2epsilon_0) [sqrt(a^2+H^2)-H]+mgH` ..............ii
At equlibrium position, `F=(-dU)/(dH)=0`
`Differentiating Eqn ii w.r.t H,
or `mg+(sigmaq)/(2epsilon_0)[(1/2)(2H)=1/sqrt(a^2+H^2)-1]=0` `[(sigmaq)/(2epsilon_0)=2mg)`
`mg+2mg[H/sqrt(a^2+H^2)-1]=0`
or `1+(2H)/sqrt(a^2+H^2)-2=0implies(2H)/sqrt(a^2+H^2)=1`
or `H=a/sqrt3`
From eqn ii we can see that
`U+2 mga` at `H=0` and
`U=U_min=sqrtmga` ag `H=a/sqrt y3`
Therefore U-H graph wil bbe as shown.
Note that at `H=a/sqrt3,U` is minimum.
`Therefore, `H=a/sqrt3` is stable equlibrium position.