A charged particle of mass m = 2 kg and charge `1 muC` is projected from a horizontal ground at an angle `theta=45^@` with speed `10ms^(-1)`. In space, a horizontal electric field towards the direction of projection `E = 2 xx 10^(7) NC^(-1)` exists. The range of the projectile is

Options

A. 20 m
B. 60 m
C. 200 m
D. 180 m

Explanation

To solve the problem, we need to find the range of a charged particle projected in the presence of an electric field. Here’s a step-by-step solution:

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

### Step 2: Calculate the time of flight
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting the values:
\[ T = \frac{2 \times 10 \times \sin(45^\circ)}{10} \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ T = \frac{2 \times 10 \times \frac{1}{\sqrt{2}}}{10} = \frac{2}{\sqrt{2}} = \sqrt{2} \, \text{s} \]

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

### Step 4: Calculate the horizontal component of the initial velocity
The horizontal component of the initial velocity \( u_x \) is given by:
\[ u_x = u \cos \theta \]
Substituting the values:
\[ u_x = 10 \cos(45^\circ) = 10 \times \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, m/s \]

### 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 values:
\[ R = \left(\frac{10}{\sqrt{2}}\right) \cdot \sqrt{2} + \frac{1}{2} \cdot 10 \cdot (\sqrt{2})^2 \]
This simplifies to:
\[ R = 10 + \frac{1}{2} \cdot 10 \cdot 2 = 10 + 10 = 20 \, m \]

### Final Answer
The range of the projectile is \( R = 20 \, m \).

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