A charged particle of mass `m=1` kg and charge `q=2muC` is thrown for a horizotal ground at an angle `theta=45^@` with speed `20 m//s`. In space a horizontal electric field `E=2xx10^(7)` V/m exist. Find the range on horizontal ground of the projectile thrown.

Explanation

To find the range of the charged particle thrown at an angle of \( \theta = 45^\circ \) with an initial speed of \( 20 \, \text{m/s} \) in the presence of a horizontal electric field \( E = 2 \times 10^7 \, \text{V/m} \), we will follow these steps:

### Step 1: Determine the initial velocity components
The initial velocity \( V \) can be resolved into horizontal and vertical components using the angle \( \theta \):
- \( V_x = V \cos(\theta) \)
- \( V_y = V \sin(\theta) \)

Given \( V = 20 \, \text{m/s} \) and \( \theta = 45^\circ \):
\[ V_x = 20 \cos(45^\circ) = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]
\[ V_y = 20 \sin(45^\circ) = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]

### Step 2: Calculate the acceleration due to the electric field
The force on the charged particle in the electric field is given by:
\[ F = qE \]
where \( q = 2 \, \mu C = 2 \times 10^{-6} \, C \) and \( E = 2 \times 10^7 \, \text{V/m} \).
\[ F = (2 \times 10^{-6}) \times (2 \times 10^7) = 4 \times 10^1 = 40 \, \text{N} \]

The acceleration \( a_x \) in the horizontal direction can be found using Newton's second law:
\[ a_x = \frac{F}{m} = \frac{40}{1} = 40 \, \text{m/s}^2 \]

### Step 3: Calculate the time of flight
The time of flight \( t \) can be calculated using the formula:
\[ t = \frac{2V_y}{g} \]
Assuming \( g \approx 10 \, \text{m/s}^2 \):
\[ t = \frac{2 \times 10\sqrt{2}}{10} = 2\sqrt{2} \, \text{s} \]

### Step 4: Calculate the range
The range \( R \) can be calculated using the formula:
\[ R = V_x t + \frac{1}{2} a_x t^2 \]
Substituting the values:
\[ R = (10\sqrt{2})(2\sqrt{2}) + \frac{1}{2} (40)(2\sqrt{2})^2 \]
Calculating each term:
1. \( (10\sqrt{2})(2\sqrt{2}) = 20 \times 2 = 40 \, \text{m} \)
2. \( (2\sqrt{2})^2 = 8 \) and thus \( \frac{1}{2} (40)(8) = 160 \, \text{m} \)

Combining both terms:
\[ R = 40 + 160 = 200 \, \text{m} \]

### Final Answer
The range on the horizontal ground of the projectile thrown is \( R = 200 \, \text{m} \).

---