Question

The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. Given the oil drop to be 1.00 um in radius and have a density of 920 kg/m^3.

a. Find the weight of the drop.
b. If the drop has a single excess electron, find the electric field strength needed to balance its weight.

Answer 1

Answer:

a) $W=3.78\cdot 10^{-14}\: N$

b) $E=2.36\cdot 10^{5} N/C$

Explanation:

a) The weight is just the mass of the drop times the acceleration of gravity.

$W=mg$

Or in terms of density.

$W=\rho Vg$ (1)

If we consider the volume of a drop as spherical, we can find the volume (R=1 μm = 0.000001 m).

$V=\frac{4}{3}\pi R^{3}$

$V=\frac{4}{3}\pi (10^{-6})^{3}$  

$V=\frac{4}{3}\pi 10^{-18}$  

$V=4.19\cdot 10^{-18}\: m^{3}$

Then, using equation (1), the weight will be:

$W=920\cdot 4.19\cdot 10^{-18}\cdot 9.81$

$W=3.78\cdot 10^{-14}\: N$  

b) To balance the weight, the electric field times the charge must be the same value in the opposite direction, which means:

$W-qE=0$                  

$E=\frac{W}{q_{electron}}$

$E=\frac{3.78\cdot 10^{-14}}{1.60\cdot 10^{-19}}$        

$E=2.36\cdot 10^{5} N/C$          

I hope it helps you!