The bob of a simple pendulum has mass `2g` and a charge of `5.0muC`. It is at rest in a uniform horizontal electric field of intensity `2000V//m`. At equilibrium, the angle that the pendulum makes with the vertical is: (take `g=10m//s^(2)`)

Options

A. `tan^(-1)(2.0)`
B. `tan^(-1)(0.2)`
C. `tan^(-1)(5.0)`
D. `tan^(-1)(0.5)`

Answer

To solve the problem of finding the angle that the pendulum makes with the vertical when it is in equilibrium in a uniform horizontal electric field, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Forces Acting on the Bob:**
- The forces acting on the bob are:
- The gravitational force (weight) acting downwards, \( F_g = mg \).
- The electric force acting horizontally due to the electric field, \( F_e = qE \).

2. **Calculate the Gravitational Force:**
- The mass of the bob is given as \( 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} \).
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).
- Therefore, the gravitational force is:
\[ F_g = mg = (2 \times 10^{-3} \, \text{kg})(10 \, \text{m/s}^2) = 2 \times 10^{-2} \, \text{N} = 0.02 \, \text{N} \]

3. **Calculate the Electric Force:**
- The charge on the bob is given as \( q = 5.0 \, \mu\text{C} = 5.0 \times 10^{-6} \, \text{C} \).
- The electric field intensity is \( E = 2000 \, \text{V/m} \).
- Therefore, the electric force is:
\[ F_e = qE = (5.0 \times 10^{-6} \, \text{C})(2000 \, \text{V/m}) = 1.0 \times 10^{-3} \, \text{N} = 0.001 \, \text{N} \]

4. **Set Up the Equilibrium Condition:**
- At equilibrium, the horizontal electric force \( F_e \) is balanced by the horizontal component of the tension \( T \) in the string, and the vertical component of the tension balances the weight of the bob.
- Let \( T \) be the tension in the string and \( \theta \) be the angle with the vertical.
- The components of the tension can be expressed as:
- Vertical component: \( T \cos(\theta) = mg \)
- Horizontal component: \( T \sin(\theta) = F_e \)

5. **Divide the Two Equations:**
- Dividing the horizontal and vertical components gives:
\[ \frac{T \sin(\theta)}{T \cos(\theta)} = \frac{F_e}{mg} \]
- This simplifies to:
\[ \tan(\theta) = \frac{F_e}{mg} \]

6. **Substitute the Values:**
- Substitute \( F_e = 0.001 \, \text{N} \) and \( mg = 0.02 \, \text{N} \):
\[ \tan(\theta) = \frac{0.001}{0.02} = \frac{1}{20} \]

7. **Calculate the Angle \( \theta \):**
- To find \( \theta \), take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{1}{20}\right) \]

### Final Answer:
The angle \( \theta \) that the pendulum makes with the vertical is:
\[ \theta \approx 2.86^\circ \]