Question

Millikan measured the electron's charge by observing tiny charged oil drops in an electric field. Each drop had a charge imbalance of only a few electrons. The strength of the electric field was adjusted so that the electric and gravitational forces on a drop would balance and the drop would be suspended in air. In this way the charge on the drop could be calculated. The charge was always found to be a small multiple of 1.60 × 10-19 C. Find the charge on an oil drop weighing 1.00 × 10-14 N and suspended in a downward field of magnitude 2.08 × 104 N/C.

Answer 1

Answer:

$4.8\cdot 10^{-19} C$

Explanation:

For a drop in equilibrium, the weight is equal to the electric force (in magnitude):

$W = F_e$

where here we have

$W=1.00\cdot 10^{-14}N$ is the weight of the drop

$F_e$ is the magnitude of the electric force, which can be rewritten as

$F_e = qE$

where

q is the charge of the oil drop

$E=2.08 \cdot 10^4 N/C$ is the magnitude of the electric field

Substituting into the equation and solving for q, we find the charge of the oil drop:

$q=\frac{W}{F_e}=\frac{1.00\cdot 10^{-14}N}{2.08\cdot 10^4 N/C}=4.8\cdot 10^{-19} C$