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 2.5 x 10⁻¹⁴ N, 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

Answer:

156250 N/C

Explanation:

weight of drop, W = 2.5 x 10^-14 N

charge of drop, q = 1.6 x 10^-19 C

Let the electric field is E.

Weight is balanced by the electrostatic force

W = qE

2.5 x 10^-14 = 1.6 x 10^-19 x E

E = 156250 N/C

Answer 2

Answer:

$1.56\times 10^5 N/C$        

Explanation:

The electric field that will balance the weight of the oil drop can be calculated using the following:

electric force, F = e E ( where, e is the charge of an electron and E is the electric field)

weight, W = 2.5 ×10⁻¹⁴ N

e E = W

$E =\frac{W}{e}$

Substitute the values:

$E =\frac{ 2.5 \times 10^{-14} N}{1.6\times10^{-19}C}= 1.56\times 10^5 N/C$