The bob of a pendulum of mass `8 mu g` carries an electric charge of `39.2xx10^(-10)` coulomb in an electric field of `20xx10^(3)` volt/meter and it is at rest. The angle made by the pendulum with the vertical will be

Options

A. `27^(@)`
B. `45^(@)`
C. `87^(@)`
D. `127^(@)`

Answer

To solve the problem step by step, we will analyze the forces acting on the pendulum bob and use trigonometric relationships to find the angle made by the pendulum with the vertical.

### Step 1: Identify the Forces Acting on the Pendulum Bob
The forces acting on the pendulum bob are:
1. The gravitational force (weight) acting downward: \( F_g = mg \)
2. The electric force acting horizontally due to the electric field: \( F_e = QE \)

Where:
- \( m = 8 \, \mu g = 8 \times 10^{-6} \, kg \)
- \( Q = 39.2 \times 10^{-10} \, C \)
- \( E = 20 \times 10^{3} \, V/m \)

### Step 2: Write the Expressions for the Forces
The weight of the bob can be calculated as:
\[ F_g = mg = (8 \times 10^{-6} \, kg)(9.8 \, m/s^2) = 7.84 \times 10^{-5} \, N \]

The electric force acting on the bob is given by:
\[ F_e = QE = (39.2 \times 10^{-10} \, C)(20 \times 10^{3} \, V/m) = 7.84 \times 10^{-6} \, N \]

### Step 3: Set Up the Equilibrium Conditions
At equilibrium, the forces acting on the bob must balance. The tension \( T \) in the string has two components:
- \( T \cos \theta = F_g \) (balances the weight)
- \( T \sin \theta = F_e \) (balances the electric force)

### Step 4: Divide the Two Equations
Dividing the two equations gives:
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{F_e}{F_g} \]
This simplifies to:
\[ \tan \theta = \frac{F_e}{F_g} \]

### Step 5: Substitute the Values
Substituting the values we calculated:
\[ \tan \theta = \frac{7.84 \times 10^{-6} \, N}{7.84 \times 10^{-5} \, N} = \frac{1}{10} \]

### Step 6: Calculate the Angle
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{1}{10}\right) \]

Using a calculator, we find:
\[ \theta \approx 5.71^\circ \]

### Conclusion
The angle made by the pendulum with the vertical is approximately \( 5.71^\circ \).

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