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Q.

A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength 81π7×105Vm1. When the field is switched off, the drop is observed to fall with terminal velocity 2×103ms1. Given g=9.8ms2, viscosity of the air =1.8×105Nsm2 and the density of oil =900 kg m1, the magnitude of q is x x 10-19C. The value of x is___

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answer is 8.

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Detailed Solution

When the electric field is on 

In equilibrium, force due to electric field = weight qE=mg

qE=43πR3ρg q=4πR3ρg3E     ...(i) When the electric field is switched off

Weight of the drop= viscous & force on the drop mg=6πηRv1

43πR3ρg=6πηR1 R=9ηv12ρg  ..(ii)

from eq. (i) & (ii) q=43π9ηv12ρg32×ρgE

=43×π9×1.8×105×2×1032×900×9.832×900×9.8×781π×105 or ,q=7.8×1019C8x10-19C

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