24i am not solving the question........let others try it.........
here is expression for tapan........
A non conducting disc of radius a and uniform positive surface charge density σ is placed on the ground with its axis vertical. A particle of mass m and +ve charge q is dropped along the axis of the disc from a height H with zero initial velocity. The particle has q/m=4ε0g/σ
a) find the value of H if the particle just reaches the disc
b) sketch the potential energy of the particle as a function of the height and find the equilibrium position.
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6 Answers
21is dis rite.....
Find Potential at ht. H
Find PE at H.......
Now equate PE at centre [ [7] ] of disc coz its JUS...... + KE gained due to FREE fALL
and equate it to PE at H
24
106Potential energy at height H = σ/2ε0*(√a2+H2-H) [formula by eureka]
PE at centre of disc = σ/2ε0*a - mgH
ΔPE = 0
==> σ/2ε0*(√a2+H2-H) = σ/2ε0*a - mgH
==> H = σa(σ/2ε0 - mg)/mg(σ-mgε0)
PE of particle at any height h < H = σ/2ε0*(√a2+h2-h) - mg(H-h)
21wud this work ??
PE at pt. height H above disc + mgH = PE at center of DISC
106ya that's what u can rearrange in my equation too.. depends on where u take grav PE = zero