Question

use the following information to answer question 4.
a simplified version of millikan’s experiment involves adjusting the potential difference between two parallel plates and observing the motion a charged oil drop inside the uniform electric field. the forces acting on the drop may be analyzed to determine the charge on the drop.

1. a 4.84 x 10⁻¹⁴ kg oil drop is suspended between two horizontal parallel plates where the electric field strength is 1.10 x 10⁴ n/c. determine the magnitude of the charge on the oil drop. 4.32 x 10⁻¹⁷ c

Explanation:

Step1: Identify Forces

The oil drop is suspended, so electric force \( F_E \) equals gravitational force \( F_g \). \( F_g = mg \), \( F_E = qE \). Thus, \( qE = mg \).

Step2: Solve for Charge \( q \)

Rearrange formula: \( q = \frac{mg}{E} \).
Given \( m = 4.84 \times 10^{-14}\, \text{kg} \), \( g = 9.8\, \text{m/s}^2 \), \( E = 1.10 \times 10^4\, \text{N/C} \).

Calculate \( mg \):
\( mg = (4.84 \times 10^{-14}\, \text{kg})(9.8\, \text{m/s}^2) = 4.7432 \times 10^{-13}\, \text{N} \).

Divide by \( E \):
\( q = \frac{4.7432 \times 10^{-13}\, \text{N}}{1.10 \times 10^4\, \text{N/C}} \approx 4.312 \times 10^{-17}\, \text{C} \) (matches given [4.32×10⁻¹⁷ C] due to rounding).

Answer:

The magnitude of the charge on the oil drop is approximately \( \boldsymbol{4.32 \times 10^{-17}\, \text{C}} \).