Question

A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field of ${100 Vm^{-1}}$. If the mass of the drop is ${1.6 × 10^{‒3}g}$, the number of electrons carried by the drop is ${(g = 10ms^{‒2})}$
A. ${10^8}$
B. ${10^{15}}$
C. ${10^{12}}$
D. ${10^6}$

Answer 1

Hint: Here the oil drop is negatively charged and Electric field is applied to overcome the force of gravitation. As the electric field applies force on a charged body, it will apply the force on the oil drop in the opposite direction as that of the electric field. So equate both the forces and get the answer.



Formula used:
The formula used in the given question is-
$n\times{e}\times{E}=m\times{g}$
Where; n = number of electrons
e = charge on one electron
E = electric field applied
m = mass of the drop
g = acceleration due to gravity








Complete answer:

The force applied by the electric field on the oil drop is-
F=$\ q\times{E}$
Since, q=$n\times{e}$
F=$\ n\times{e}\times{E}$ ------------(i)
The force applied by the gravitational force on the oil drop is-
F=$\ m\times{g}$ ------------(ii)
Equating equation (i) and (ii)
$n\times{e}\times{E}=m\times{g}$
Given;
e = ${1.6 × 10^{‒19} C}$
E = ${100 Vm^{‒1}}$
m = ${1.6 × 10^{‒3}g}$ = ${1.6 × 10^{‒6} Kg}$
g = ${10ms^{‒2}}$
putting all the values in the equation,
$n\times1.6\ \times\ {10}^{-19}\times100=\ 1.6\ \times\ {10}^{-6}\times10$
 $n=\frac{1.6\ \times\ {10}^{-6}\times10}{1.6\ \times\ {10}^{-19}\times100}$
$n=\frac{\ {10}^{-5}}{\ {10}^{-17}}$
$n={10}^{-5+17}$
$n={10}^{12}$
Thus the number of electrons carried by the drop is ${10^{12}}$ electrons.
Thus, Option (C) is correct



Note: Electric field is a vector quantity. If the charge is negative the electric field and the force will be in the opposite direction and if the charge is positive the electric field and the force will be in the same direction.