A charged particle of mass `m = 1` kg and charge `q = 2muC` is thrown from a horizontal ground at an angle `theta = 45^@` with the speed `25 m//s`. In space, a horizontal electric field `E = 2 xx 10^7` V/m exists in the direction of motion. Find the range on horizontal ground of the projectile thrown. Take `g = 10 m//s^2`.

Explanation

To find the range of the charged particle thrown at an angle of 45 degrees in the presence of an electric field, we will follow these steps:

### Step 1: Identify the given values
- Mass of the particle, \( m = 1 \, \text{kg} \)
- Charge of the particle, \( q = 2 \, \mu\text{C} = 2 \times 10^{-6} \, \text{C} \)
- Initial speed, \( u = 25 \, \text{m/s} \)
- Angle of projection, \( \theta = 45^\circ \)
- Electric field, \( E = 2 \times 10^7 \, \text{V/m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the horizontal and vertical components of the initial velocity
The initial velocity can be resolved into horizontal and vertical components using trigonometric functions:
- Horizontal component, \( u_x = u \cos \theta = 25 \cos 45^\circ = 25 \times \frac{1}{\sqrt{2}} = \frac{25}{\sqrt{2}} \approx 17.68 \, \text{m/s} \)
- Vertical component, \( u_y = u \sin \theta = 25 \sin 45^\circ = 25 \times \frac{1}{\sqrt{2}} = \frac{25}{\sqrt{2}} \approx 17.68 \, \text{m/s} \)

### Step 3: Calculate the horizontal acceleration due to the electric field
The horizontal acceleration \( a_x \) caused by the electric field can be calculated using the formula:
\[ a_x = \frac{qE}{m} \]
Substituting the values:
\[ a_x = \frac{(2 \times 10^{-6}) \times (2 \times 10^7)}{1} = 0.04 \, \text{m/s}^2 \]

### Step 4: Calculate the total time of flight
The total time of flight \( T \) can be calculated using the formula:
\[ T = \frac{2u_y}{g} \]
Substituting the values:
\[ T = \frac{2 \times 17.68}{10} = 3.536 \, \text{s} \]

### Step 5: Calculate the range
The range \( R \) can be calculated using the formula:
\[ R = u_x T + \frac{1}{2} a_x T^2 \]
Substituting the known values:
\[ R = (17.68)(3.536) + \frac{1}{2}(0.04)(3.536)^2 \]
Calculating each term:
1. \( u_x T = 17.68 \times 3.536 \approx 62.5 \, \text{m} \)
2. \( \frac{1}{2} a_x T^2 = \frac{1}{2} \times 0.04 \times (3.536)^2 \approx 0.5 \times 0.04 \times 12.5 \approx 0.25 \, \text{m} \)

Adding these together:
\[ R \approx 62.5 + 0.25 = 62.75 \, \text{m} \]

### Final Result
The range of the projectile on the horizontal ground is approximately \( R \approx 62.75 \, \text{m} \).

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