Question

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 1.6X10^-3g, the number of electrons carried by the drop is

Answer 1

9.8 *10^11 electrons

Answer 2

Answer:

The number of electrons carried by the drop is = $10^{15}$

Explanation:

Downward force due to gravity,

Force due to gravity is given by, $F_g = ma$

where, $m$ is the mass of the oil drop,

$a$ is the acceleration due to gravity = $10m/s^2$.

The upward force is due to electric field,

Force due to electric field on a charge is given by, $F_E=qE$

where, $q$ is the charge of the oil drop,

$E$ is the magnitude of electric field.

To prevent oil drop from falling, Upward and downward force must be equal.

Upward force = downward force

$\implies qE=mg$

$\implies q=\frac{mg}{E}$

Putting in the values,

$\implies q=\frac{1.6*10^{-3}*10}{100}$

$\implies q=1.6*10^{-4}C$

Number of electrons which constitute this charge = $\frac{q}{1.6*10^{-19}}$

where, $1.6*10^{-19}$C is the charge of 1 electron.

Number of electrons = $\frac{1.6*10^{-4}}{1.6*10^{-19}} = 10^{15}$

The number of electrons carried by the drop is = $10^{15}$

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