A `1kg` ball is suspended in a uniform electric field with the help of a string fixed to a point. The ball is given a charge `sqrt(5)` coulomb and the string makes an angle `37^(@)` with the vertical in equilibrium position. In the equilibrium position the tension is double the weight of the ball. Find the magnitude of the electric field in `N//C`.

Explanation

To solve the problem, we need to analyze the forces acting on the charged ball in equilibrium and use the given information to find the magnitude of the electric field.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Ball:**
- The weight of the ball (W) acting downwards: \( W = mg \)
- The tension (T) in the string acting at an angle of \( 37^\circ \) with the vertical.
- The electric force (F_e) acting horizontally due to the electric field.

2. **Given Information:**
- Mass of the ball (m) = 1 kg
- Charge of the ball (Q) = \( \sqrt{5} \) C
- Angle with the vertical = \( 37^\circ \)
- Tension (T) is double the weight of the ball: \( T = 2W = 2mg \)

3. **Calculate the Weight of the Ball:**
\[ W = mg = 1 \, \text{kg} \times 10 \, \text{m/s}^2 = 10 \, \text{N} \]

4. **Calculate the Tension in the String:**
\[ T = 2W = 2 \times 10 \, \text{N} = 20 \, \text{N} \]

5. **Resolve the Tension into Components:**
- Vertical component of tension (T_y):
\[ T_y = T \cos(37^\circ) = 20 \cos(37^\circ) \]
- Horizontal component of tension (T_x):
\[ T_x = T \sin(37^\circ) = 20 \sin(37^\circ) \]

6. **Using the Equilibrium Condition:**
- In the vertical direction:
\[ T_y = W \implies 20 \cos(37^\circ) = 10 \]
- In the horizontal direction, the electric force (F_e) is equal to the horizontal component of tension:
\[ F_e = T_x = 20 \sin(37^\circ) \]

7. **Calculate the Electric Force:**
- Using \( \sin(37^\circ) \approx 0.6 \):
\[ F_e = 20 \times 0.6 = 12 \, \text{N} \]

8. **Relate Electric Force to Electric Field:**
- The electric force is given by:
\[ F_e = QE \implies E = \frac{F_e}{Q} \]
- Substituting the values:
\[ E = \frac{12 \, \text{N}}{\sqrt{5} \, \text{C}} \]

9. **Calculate the Magnitude of the Electric Field:**
\[ E = \frac{12}{\sqrt{5}} \approx \frac{12}{2.236} \approx 5.37 \, \text{N/C} \]

### Final Answer:
The magnitude of the electric field is approximately \( 5.37 \, \text{N/C} \).