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Question

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 charge q is dropped, along the axis of the disc from a height H with zero initial velocity. It was observed that its velocity becomes zero just when it reaches the disc. The value of 3Ha is (​​​​​​​Given qm=4ϵogσ)

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Solution


Since the initial and final velocities are both zero, we can say

Δ K.E = 0

Work done = 0

i.e Work done by gravity (Wg) + Work done by electric field (We) = 0

i.e mgH+We=0

Work done by an electric field in moving charge from point A to point B =q(VBVA)

i.e mgH+(q(VoVP))=0

i.e mgH=q(VoVP)

Potential due to a disc of radius a at a point at a distance x on its axis

V=σ2ϵ0(a2+x2x)

(see video for derivation)

Vo=σ2ϵ0a and
VP=σ2ϵ0(a2+H2H)

Substituting in the work energy equation,

mgH+qσ2ϵ0(aa2+H2+H)

gH=qσ2ϵ0m(aa2+H2+H)

gH=2g(aa2+H2+H)

(Given qm=4ϵoga)

H2=(a+Ha2+H2)

a2+H2=a+H2

a2+H2=a2+H24+aH

3H24=aH

3Ha=4

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