To solve the problem, we need to find the number of electrons that must be attached to an oil drop so that it moves with a constant speed between two parallel conducting plates. Here’s a step-by-step solution:
### Step 1: Understand the Forces Acting on the Oil Drop
When the oil drop is between the two plates, two main forces act on it:
- The gravitational force (weight) acting downward, given by \( F_g = mg \).
- The electric force acting upward due to the electric field between the plates, given by \( F_e = QE \), where \( Q \) is the charge on the oil drop and \( E \) is the electric field strength.
### Step 2: Calculate the Electric Field Strength
The electric field \( E \) between two parallel plates is given by the formula:
\[
E = \frac{V}{d}
\]
where:
- \( V = 1.5 \, \text{kV} = 1.5 \times 10^3 \, \text{V} \)
- \( d = 1.5 \times 10^{-2} \, \text{m} \)
Substituting the values:
\[
E = \frac{1.5 \times 10^3}{1.5 \times 10^{-2}} = 1.0 \times 10^5 \, \text{V/m}
\]
### Step 3: Set Up the Equation for Constant Speed
For the oil drop to move with constant speed, the net force acting on it must be zero. Therefore, the gravitational force must equal the electric force:
\[
mg = QE
\]
Here, \( Q \) can be expressed in terms of the number of electrons \( n \):
\[
Q = n \cdot e
\]
where \( e = 1.6 \times 10^{-19} \, \text{C} \) (the charge of a single electron).
### Step 4: Substitute and Rearrange the Equation
Substituting \( Q \) into the force balance equation gives:
\[
mg = n \cdot e \cdot E
\]
Rearranging for \( n \):
\[
n = \frac{mg}{eE}
\]
### Step 5: Plug in the Values
We know:
- \( m = 4.8 \times 10^{-15} \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( E = 1.0 \times 10^5 \, \text{V/m} \)
Now substituting these values into the equation:
\[
n = \frac{(4.8 \times 10^{-15} \, \text{kg})(10 \, \text{m/s}^2)}{(1.6 \times 10^{-19} \, \text{C})(1.0 \times 10^5 \, \text{V/m})}
\]
Calculating the numerator:
\[
4.8 \times 10^{-15} \times 10 = 4.8 \times 10^{-14} \, \text{kg m/s}^2
\]
Calculating the denominator:
\[
1.6 \times 10^{-19} \times 1.0 \times 10^5 = 1.6 \times 10^{-14} \, \text{C V/m}
\]
Now substituting these into the equation for \( n \):
\[
n = \frac{4.8 \times 10^{-14}}{1.6 \times 10^{-14}} = 3
\]
### Conclusion
The number of electrons that must be attached to the oil drop is \( n = 3 \).
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