A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 text Vm-1 . If the mass of the drop is 1.6× 10-3g, the number of electrons carried by the drop is (g=10 text ms-2)

Question

A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 text Vm-1 . If the mass of the drop is 1.6× 10-3g, the number of electrons carried by the drop is (g=10 text ms-2)  (A)  1018  (B)  1015  (C)  106  (D)  109

Tardigrade Admin · Accepted Answer

qE=mg or ne× E=mg n× 1.6× 10-19× 100=1.6× 10-6× 10 n=1012

Q. A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field $100 V m^{- 1}$ . If the mass of the drop is $1.6 \times 10^{- 3} g,$ the number of electrons carried by the drop is $(g = 10 m s^{- 2})$

$10^{18}$

$10^{15}$

$10^{6}$

$10^{9}$

$10^{12}$

$qE = m g$
or $n e \times E = m g$
$n \times 1.6 \times 10^{- 19} \times 100 = 1.6 \times 10^{- 6} \times 10$
$n = 10^{12}$

Q. A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100Vm−1 . If the mass of the drop is 1.6×10−3g, the number of electrons carried by the drop is (g=10ms−2)

1018

1015

106

109

1012

qE=mg or ne×E=mg n×1.6×10−19×100=1.6×10−6×10 n=1012