A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of stength 81pie/7 *10^5 V/m. when the field is switched off , the drop is observed to fall wth terminal velocity 2*10^-3m/s. given g=9.8m/s^2, viscocity of the air = 1.8*10^-5Ns/m^2 and the density of the oil =900kg/m^3. the magnitude of q=? A)1.6*10^-19C B)3.2*10^-19C C)4.8*10^-19C D)8.0*10^-19C
\[qE=4/3 \pi*r^3*\rho*g\]
when the drop descends with constant velocity v in the absence of electric field, then mg = \[4/3π∗r3∗ρ∗g=6*\pi *\eta*r*v\]
from here \[r^2=(9/2) *(v*\eta/\rho*g)\]
from here calculate r =3/7 * 10^-5
\[6∗π∗η∗r∗v = qE \]
\[(6∗π∗η∗r∗v)/E=q\]
=8.0*10^-19C
ab values dal kar calculate kar lena ans yahi aye ga may be mera to yahi aya hai
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