Millikan's oil drop

Electric Fields - OCR A-Level Physics

Worked Example
An oil drop of mass 9.79 × 10⁻¹⁵ kg is held stationary between parallel plates separated by 5.0 mm with a p.d. of 306 V. Calculate the chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). on the drop and the number of excess electrons.
Show Solution
1

The drop is stationary so the electric force balances gravity: $QE = mg$.

2
Electric field between plates

$E =\;\text{V}/d = 306 / (5.0 \times 10⁻^{3}) = 6.12 \times 10⁴\;\text{V}\;\text{m}⁻¹.$

3
Rearrange for Q

$Q =\;\text{m}g/E = (9.79 \times 10⁻¹⁵)(9.81) / (6.12 \times 10⁴).$

4

$Q = 9.60 \times 10⁻¹⁴ / 6.12 \times 10⁴ = 1.57 \times 10⁻¹⁸ C.$

5
Number of electrons

$n = Q/e = 1.57 \times 10⁻¹⁸ / 1.60 \times 10⁻¹⁹ ≈ 9.8 ≈ 10 electrons.$

6

Since chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). is quantisedExisting only in discrete values. ChargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). is quantised because it can only exist as whole-number multiples of the elementary charge e., n must be a whole number: $n = 10$.

Answer
Q = 1.60 × 10⁻¹⁸ C (10 elementary charges). The drop carries 10 excess electrons.
Electric Fields Overview
Coulomb's law 10 of 10