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 solve the problem of finding the range of a charged particle thrown at an angle with a horizontal electric field present, we will follow these steps:

### Step 1: Determine the initial velocity components
The initial velocity \( u \) of the particle is given as \( 20 \, \text{m/s} \) and the angle \( \theta \) is \( 45^\circ \).

- The horizontal component of the initial velocity \( u_x \) is:
\[ u_x = u \cos \theta = 20 \cos 45^\circ = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]

- The vertical component of the initial velocity \( u_y \) is:
\[ u_y = u \sin \theta = 20 \sin 45^\circ = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]

### Step 2: Calculate the time of flight
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u_y}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

Substituting the values:
\[ T = \frac{2 \times 10\sqrt{2}}{9.81} \approx \frac{20\sqrt{2}}{9.81} \approx 2.86 \, \text{s} \]

### Step 3: Calculate the force due to the electric field
The force \( F \) acting on the charged particle in the horizontal direction due to the electric field \( E \) is given by:
\[ F = qE \]
where \( q = 2 \, \mu C = 2 \times 10^{-6} \, C \) and \( E = 2 \times 10^7 \, V/m \).

Calculating the force:
\[ F = (2 \times 10^{-6}) \times (2 \times 10^7) = 40 \, \text{N} \]

### Step 4: Calculate the horizontal acceleration
Using Newton's second law, the acceleration \( a \) in the horizontal direction is:
\[ a = \frac{F}{m} \]
where \( m = 1 \, \text{kg} \).

Substituting the values:
\[ a = \frac{40}{1} = 40 \, \text{m/s}^2 \]

### Step 5: Calculate the horizontal range
The range \( R \) can be calculated using the formula:
\[ R = u_x T + \frac{1}{2} a T^2 \]

Substituting the values:
\[ R = (10\sqrt{2})(2.86) + \frac{1}{2}(40)(2.86)^2 \]
Calculating \( R \):
1. Calculate \( 10\sqrt{2} \times 2.86 \):
\[ 10\sqrt{2} \approx 14.14 \quad \Rightarrow \quad 14.14 \times 2.86 \approx 40.4 \]
2. Calculate \( \frac{1}{2} \times 40 \times (2.86)^2 \):
\[ (2.86)^2 \approx 8.18 \quad \Rightarrow \quad \frac{1}{2} \times 40 \times 8.18 \approx 163.6 \]

Adding both parts together:
\[ R \approx 40.4 + 163.6 = 204 \, \text{m} \]

### Final Answer
The range of the projectile on the horizontal ground is approximately \( 204 \, \text{m} \).

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