To tackle the problem presented in the Millikan oil drop experiment, we need to break it down into manageable parts. The experiment's goal is to measure the charge of an electron by observing the behavior of charged oil droplets in an electric field. Let's go through the calculations step by step.
Step A: Finding the Radius of the Drop
First, we need to determine the radius of the oil drop. We can use the formula for the volume of a sphere and the mass density to find the radius. The volume \( V \) of a sphere is given by:
V = \frac{4}{3} \pi r^3
Where \( r \) is the radius of the drop. The mass \( m \) of the drop can be expressed as:
m = \text{density} \times V = \text{density} \times \frac{4}{3} \pi r^3
Given that the density of the oil drop is 880 kg/m³, we can express the mass as:
m = 880 \times \frac{4}{3} \pi r^3
Next, we need to relate the mass to the gravitational force acting on the drop. The gravitational force \( F_g \) can be calculated using:
F_g = m \cdot g
Where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). When the electric field is switched off, the drop falls under gravity, and we can use the distance fallen to find the radius. The distance \( d \) fallen in time \( t \) can be described by the equation:
d = \frac{1}{2} g t^2
Substituting the values \( d = 0.002 \) m (2.0 mm) and \( t = 35.7 \) s, we can find \( g \):
0.002 = \frac{1}{2} \cdot 9.81 \cdot (35.7)^2
Calculating gives us:
0.002 = \frac{1}{2} \cdot 9.81 \cdot 1276.49
Now, solving for \( r \) involves rearranging the mass equation and substituting the known values. However, we can also find the terminal velocity \( v_t \) when the drop is falling freely under gravity:
v_t = \frac{d}{t} = \frac{0.002}{35.7} \approx 5.6 \times 10^{-5} \text{ m/s}
At terminal velocity, the gravitational force equals the drag force. The drag force \( F_d \) can be expressed using Stokes' law:
F_d = 6 \pi \eta r v_t
Where \( \eta \) is the dynamic viscosity of air (approximately \( 1.81 \times 10^{-5} \) Pa·s). Setting \( F_g = F_d \):
m \cdot g = 6 \pi \eta r v_t
Substituting \( m \) and rearranging gives us:
880 \cdot \frac{4}{3} \pi r^3 \cdot 9.81 = 6 \pi (1.81 \times 10^{-5}) r (5.6 \times 10^{-5})
Solving this equation will yield the radius \( r \) of the drop.
Step B: Calculating the Charge on the Drop
Once we have the radius, we can find the charge on the drop. When the electric field is applied, the electric force \( F_e \) acting on the drop is given by:
F_e = qE
Where \( q \) is the charge on the drop and \( E \) is the electric field strength. The electric field \( E \) between the plates can be calculated using:
E = \frac{V}{d}
Substituting the potential difference \( V = 103 \) V and the distance \( d = 0.006 \) m:
E = \frac{103}{0.006} \approx 17166.67 \text{ V/m}
Now, when the electric field is on, the drop is held stationary, meaning the electric force equals the gravitational force:
qE = mg
Substituting \( m \) and rearranging gives us:
q = \frac{mg}{E}
Using the mass we calculated earlier and the electric field strength, we can find the charge \( q \) on the drop.
By following these steps, you can estimate both the radius of the oil drop and the charge it carries. This experiment beautifully illustrates the principles of electrostatics and fluid dynamics in a practical setting.
