A pendulum bob of mass `30.7 xx 10^(-6)kg` and carrying a chargee `2 xx 10^(-8)C` is at rest in a horizontal uniform electric field of `20000V//m`. The tension in the thread of the pendulum is `(g = 9.8 m//s^(2))`

Options

A. `3 xx 10^(4)N`
B. `4 xx 10^(-4)N`
C. `5 xx 10^(-4)N`
D. `6 xx 10^(-4)N`

Answer

To solve the problem, we will analyze the forces acting on the pendulum bob and use the equilibrium condition since the bob is at rest.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Pendulum Bob:**
- The weight of the bob acting downwards: \( F_g = mg \)
- The electric force acting horizontally due to the electric field: \( F_e = qE \)
- The tension in the thread acting at an angle \( \theta \) to the vertical.

2. **Write the Expressions for the Forces:**
- The weight of the bob is given by:
\[ F_g = mg \]
- The electric force is given by:
\[ F_e = qE \]

3. **Set Up the Equilibrium Conditions:**
- Since the bob is at rest, the net force in both the vertical and horizontal directions must be zero.
- In the vertical direction:
\[ T \cos \theta = mg \]
- In the horizontal direction:
\[ T \sin \theta = qE \]

4. **Square and Add the Two Equations:**
- Squaring both equations and adding them gives:
\[ (T \cos \theta)^2 + (T \sin \theta)^2 = (mg)^2 + (qE)^2 \]
- This simplifies to:
\[ T^2 (\cos^2 \theta + \sin^2 \theta) = m^2g^2 + q^2E^2 \]
- Since \( \cos^2 \theta + \sin^2 \theta = 1 \), we have:
\[ T^2 = m^2g^2 + q^2E^2 \]

5. **Calculate the Values:**
- Given:
- Mass \( m = 30.7 \times 10^{-6} \, \text{kg} \)
- Charge \( q = 2 \times 10^{-8} \, \text{C} \)
- Electric field \( E = 20000 \, \text{V/m} \)
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)

- Calculate \( mg \):
\[ mg = 30.7 \times 10^{-6} \times 9.8 = 3.00686 \times 10^{-5} \, \text{N} \]
- Calculate \( qE \):
\[ qE = 2 \times 10^{-8} \times 20000 = 4 \times 10^{-4} \, \text{N} \]

6. **Substitute Back to Find Tension \( T \):**
- Now substitute \( mg \) and \( qE \) into the tension equation:
\[ T^2 = (3.00686 \times 10^{-5})^2 + (4 \times 10^{-4})^2 \]
- Calculate \( T^2 \):
\[ T^2 = (9.036 \times 10^{-10}) + (1.6 \times 10^{-7}) = 1.60936 \times 10^{-7} \]
- Therefore, \( T \) is:
\[ T = \sqrt{1.60936 \times 10^{-7}} \approx 4.01 \times 10^{-4} \, \text{N} \]

### Final Answer:
The tension in the thread of the pendulum is approximately \( 4.01 \times 10^{-4} \, \text{N} \).