Question

Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field of 46 N/C. The density of the oil is 0.085 g/cm3 . Calculate for Millikan the charge on the droplet. Is quantization of charge obeyed within the precision of this experiment?

Answer 1

Answer:

The charge on the droplet is $3.106\times10^{-16}\ C$.

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

Electric field = 46 N/C

Density of oil = 0.085 g/cm³

We need to calculate the charge on the droplet

Using formula of force

$F= qE$

$mg=qE$

$V\times\rho\times g=qE$

$q=\dfrac{V\times\rho}{E}$

$q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}$

Put the value into the formula

$q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}$

$q=3.106\times10^{-16}\ C$

We need to calculate the quantization of charge

Using formula of quantization

$n = \dfrac{q}{e}$

Put the value into the formula

$n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}$

$n=1941.25$

Yes, quantization of charge is obeyed within experimental error.

Hence, The charge on the droplet is $3.106\times10^{-16}\ C$.

Yes, quantization of charge is obeyed within experimental error.

Answer 2

Answer:

3.11 x 10^-16 C

Explanation:

radius of drop, r = 1.6 micro metre = 1.6 x 10^-6 m

Electric field, E = 46 N/C

density of oil, d = 0.085 g/cm³ = 85 kg/m³

Let the charge on the oil drop is q.

So, the weight of the drop is balanced by the electric force on the drop.

mass of drop x g = charge of the drop x electric field

m x g = q E

Volume x density x g = q E

$\frac{4}{3} \pi\times r^{3}\times d\times g = q \times E$

$\frac{4}{3} \pi\times 1.6^{3}\times 10^{-18}\times 85\times 9.8 = q \times 46$

q = 3.11 x 10^-16 C

Thus, the charge on the oil drop is 3.11 x 10^-16 C.