To solve the problem, we will analyze the forces acting on the pendulum bob in the presence of an electric field and find the angle \( \alpha \) and the tension \( T \) in the thread of the pendulum.
### Step 1: Identify the forces acting on the pendulum bob
The forces acting on the pendulum bob are:
1. The weight of the bob, \( W = mg \), acting downwards.
2. The tension in the string, \( T \), which can be resolved into two components:
- \( T \cos \alpha \) acting vertically upwards.
- \( T \sin \alpha \) acting horizontally towards the electric field.
3. The electric force due to the electric field, \( F_e = qE \), acting horizontally in the direction of the electric field.
### Step 2: Set up the equations for equilibrium
Since the pendulum bob is at rest, the forces in both the vertical and horizontal directions must balance.
**Vertical direction:**
\[
T \cos \alpha = mg \quad \text{(1)}
\]
**Horizontal direction:**
\[
T \sin \alpha = qE \quad \text{(2)}
\]
### Step 3: Divide the equations to find \( \tan \alpha \)
Dividing equation (2) by equation (1):
\[
\frac{T \sin \alpha}{T \cos \alpha} = \frac{qE}{mg}
\]
This simplifies to:
\[
\tan \alpha = \frac{qE}{mg}
\]
### Step 4: Substitute the known values
Given:
- \( m = 80 \, \text{mg} = 80 \times 10^{-6} \, \text{kg} \)
- \( q = 2 \times 10^{-8} \, \text{C} \)
- \( E = 20000 \, \text{V/m} \)
- \( g = 9.8 \, \text{m/s}^2 \)
Substituting these values into the equation:
\[
\tan \alpha = \frac{(2 \times 10^{-8} \, \text{C})(20000 \, \text{V/m})}{(80 \times 10^{-6} \, \text{kg})(9.8 \, \text{m/s}^2)}
\]
### Step 5: Calculate \( \tan \alpha \)
Calculating the numerator:
\[
qE = (2 \times 10^{-8})(20000) = 4 \times 10^{-4} \, \text{N}
\]
Calculating the denominator:
\[
mg = (80 \times 10^{-6})(9.8) = 7.84 \times 10^{-5} \, \text{N}
\]
Now substituting these values:
\[
\tan \alpha = \frac{4 \times 10^{-4}}{7.84 \times 10^{-5}} \approx 5.10
\]
### Step 6: Find \( \alpha \)
Now, we find \( \alpha \):
\[
\alpha = \tan^{-1}(5.10) \approx 27^\circ
\]
### Step 7: Calculate the tension \( T \)
Using equation (1):
\[
T = \frac{mg}{\cos \alpha}
\]
Substituting the values:
\[
T = \frac{(80 \times 10^{-6})(9.8)}{\cos(27^\circ)}
\]
Calculating \( \cos(27^\circ) \):
\[
\cos(27^\circ) \approx 0.848
\]
Now substituting:
\[
T = \frac{(80 \times 10^{-6})(9.8)}{0.848} \approx \frac{7.84 \times 10^{-5}}{0.848} \approx 9.24 \times 10^{-5} \, \text{N} \approx 880 \, \mu\text{N}
\]
### Final Results
- The angle \( \alpha \) is approximately \( 27^\circ \).
- The tension \( T \) in the thread is approximately \( 880 \, \mu\text{N} \).
