Question

In Millikan's oil-drop experiment, one looks at a small oil drop held motionless between two plates. Take the voltage between the plates to be 2036 V, and the plate separation to be 2.08 cm. The oil drop (of density 0.81 g/cm3) has a diameter of 4.00 ✕ 10−6 m. Find the charge on the drop, in terms of electron units.

Answer 1

Answer:

The charge on the drop is $2.71\times10^{-18}\ C$

Explanation:

Given that,

Voltage = 2036 V

Distance d= 2.08 cm

Diameter $D=4.00\times10^{-6}\ m$

Density $\rho=0.81\ g/cm^3$

We need to calculate the volume of the drop

$V=\dfrac{4}{3}\pi\times r^3$

Put the value into the formula

$V=\dfrac{4}{3}\times\pi\times (\dfrac{4.00\times10^{-6}}{2})^3$

$V=3.35\times10^{-17}\ m^3$

We need to calculate the mass

Using formula of density

$\rho=\dfrac{m}{V}$

$m=\rho\times V$

$m=0.81\times1000\times3.35\times10^{-17}$

$m=2.71\times10^{-14}\ kg$

We need to calculate the electric field

Using formula of electric field

$E = \dfrac{V}{d}$

Where, V = potential difference

d = separation of plates

Put the value into the formula

$E=\dfrac{2036}{2.08\times10^{-2}}$

$E=9.788\times10^{4}\ N/C$

We need to calculate the charge

Using formula of charge

$E=\dfrac{F}{q}$

$q=\dfrac{mg}{E}$

Where, E = electric field

m = mass

g = acceleration due to gravity

Put the value into the formula

$q=\dfrac{2.71\times10^{-14}\times9.8}{9.788\times10^{4}}$

$q=2.71\times10^{-18}\ C$

Hence, The charge on the drop is $2.71\times10^{-18}\ C$