A particle of mass `10^(-3)` kg and charge 1.0 c is initially at rest At time t=0 the paritcle comes under the influence of an electric field `hat (E )(t)=E_(0) sin omega t hat(i)` where `E_(0) =1.0 N C^(-1)` and `omega =10^(3) rad s^(-1)` consider the effect of only the electrical force on the particle .Then the maximum speed in m `s^(-1)` attained by the paritcle at subsequent times is ______________

Explanation

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Identify the given parameters
- Mass of the particle, \( m = 10^{-3} \, \text{kg} \)
- Charge of the particle, \( q = 1.0 \, \text{C} \)
- Electric field, \( \hat{E}(t) = E_0 \sin(\omega t) \hat{i} \) where \( E_0 = 1.0 \, \text{N/C} \) and \( \omega = 10^3 \, \text{rad/s} \)

### Step 2: Calculate the force on the particle
The force \( F \) acting on the particle due to the electric field is given by:
\[ F = q \hat{E}(t) = q E_0 \sin(\omega t) \hat{i} \]
Substituting the values:
\[ F = 1.0 \cdot 1.0 \sin(10^3 t) \hat{i} = \sin(10^3 t) \hat{i} \, \text{N} \]

### Step 3: Relate force to acceleration
According to Newton's second law, \( F = m a \), where \( a \) is the acceleration. Therefore, we can write:
\[ m a = \sin(10^3 t) \]
Thus, the acceleration \( a \) is given by:
\[ a = \frac{F}{m} = \frac{\sin(10^3 t)}{10^{-3}} = 10^3 \sin(10^3 t) \, \text{m/s}^2 \]

### Step 4: Relate acceleration to velocity
Acceleration is the rate of change of velocity:
\[ a = \frac{dv}{dt} \]
So, we can express this as:
\[ \frac{dv}{dt} = 10^3 \sin(10^3 t) \]

### Step 5: Integrate to find velocity
To find the velocity \( v \), we integrate the acceleration:
\[ dv = 10^3 \sin(10^3 t) dt \]
Integrating both sides:
\[ v = \int 10^3 \sin(10^3 t) dt \]
The integral of \( \sin(kt) \) is \( -\frac{1}{k} \cos(kt) + C \), where \( k = 10^3 \):
\[ v = -\frac{10^3}{10^3} \cos(10^3 t) + C = -\cos(10^3 t) + C \]
Since the particle is initially at rest (\( v = 0 \) at \( t = 0 \)):
\[ 0 = -\cos(0) + C \implies C = 1 \]
Thus, the velocity becomes:
\[ v(t) = 1 - \cos(10^3 t) \]

### Step 6: Find the maximum velocity
The maximum value of \( v(t) \) occurs when \( \cos(10^3 t) \) is at its minimum value, which is -1:
\[ v_{\text{max}} = 1 - (-1) = 2 \, \text{m/s} \]

### Conclusion
The maximum speed attained by the particle is:
\[ \boxed{2 \, \text{m/s}} \]