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A non-conducting disc of positive surface charge density 's' is placed horizontally on the ground. A particle of mass 'm' is dropped along the axis of the disc from height 'H' with 0 initial velocity. The particle has q/m=pi*g/sk . Find the height H such that the particle just reaches the disc.

A non-conducting disc of positive surface charge density 's' is placed horizontally on the ground. A particle of mass 'm' is dropped along the axis of the disc from height 'H' with 0 initial velocity. The particle has q/m=pi*g/sk . Find the height H such that the particle just reaches the disc.

Grade:12

1 Answers

Saurabh Kumar
askIITians Faculty 2400 Points
9 years ago
The electric potential of the disk at points along its axis needs to be determined first. We can then equate the potential energy (gravitational plus electrical) at height H to that at the disk's center, because "just reaching the disk" means that initial and final kinetic energies are zero.

Adding (integrating) contributions to the potential from infinitesimally thin rings making up the disk gives for a point on the axes at height z:

V(z) = k ∫ s 2π r dr /sqrt(r^2 + z^2)
= ks 2π ( sqrt(R^2 + z^2) - z)

So the potential energy for the particle with charge q at height H is

U = q V(H)
= 2π q k s ( sqrt(R^2 + H^2) - H )

Therefore equalling the potential energies at initial and final positions gives

m g H + 2π q k s ( sqrt(R^2 + H^2) - H )= 2π q k s R

This we have to solve for H:
2π q k s ( sqrt(R^2 + H^2) - H ) = 2π q k s R - m g H

sqrt(R^2 + H^2) - H = R - mgH/(2π q k s )

It is given that q k s = πmg, So

sqrt(R^2 + H^2) = R + H(1-1/(2π^2) )

Squaring both sides gives

R^2 + H^2 = R^2 + 2 R H (1-1/(2π^2)) + H^2 (1-1/2π^2)^2

H^2 (1 - (1-1/2π^2)^2 )= 2 R H (1-1/(2π^2))


H = 0 or H = 2 R (1-1/(2π^2))/(1 - (1-1/2π^2)^2 )

So

H = 2R (1-1/(2π^2)) / (1/π^2 - 1/(4π^4))
= 2R (4π^4 -2π^2)/(4π^2-1)
= 4π^2 (2π^2-1)/(4π^2-1) R

which is about 19.2 R

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