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. Consider an oil drop with a weight of 2.9 x 10-14N, if the drop has a single excess electron, find the magnitude (in N/C) of the electric field needed to balance its weight. Your should round your answer to an integer, indicate only the number, do not include the unit.

Answer 1

`E=1.81* 10^5\ N/C`

Step-by-step explanation:

In Millikan oil drop experiment, oil drops were suspended against the gravitational force by a vertical electric field such that its weight is balanced by the electron force i.e.

W = qE,

W is weight, W = mg

q is charge,

E is electric field

⇒
`2.9* 10^(-14)\ N=qE`

or

`E=(2.9* 10^(-14)\ N)/(1.6* 10^(-19)\ C)\\\\E=181250\ N/C\\\\\text{or}\\\\E=1.81* 10^5\ N/C`

So, the magnitude of the electric field is
`1.81* 10^5\ N/C`.