A stream of a positively charged particles `q/m=2 times 10^(11)C/kg` and velocity `vecv_theta=3 times 10^(7)hati m/s` is deflected by an electric field `1.8vecjkV/m`.Theelectric field exists in a region of 10 cm along x direction. Due to the electric field, the deflection of the charge particles in the y direction is _____mm.

Answer

To solve the problem, we need to find the deflection of positively charged particles in the y-direction due to an electric field. Let's break down the solution step by step.

### Step 1: Identify the Given Values
- Charge-to-mass ratio: \( \frac{q}{m} = 2 \times 10^{11} \, \text{C/kg} \)
- Velocity of particles: \( \vec{v} = 3 \times 10^{7} \hat{i} \, \text{m/s} \)
- Electric field: \( \vec{E} = 1.8 \hat{j} \, \text{kV/m} = 1.8 \times 10^{3} \hat{j} \, \text{V/m} \)
- Length of the region with electric field: \( x = 10 \, \text{cm} = 0.1 \, \text{m} \)

### Step 2: Calculate the Time of Flight
The time \( t \) taken to travel through the electric field can be calculated using the formula:
\[ t = \frac{x}{v_x} \]
Substituting the values:
\[ t = \frac{0.1 \, \text{m}}{3 \times 10^{7} \, \text{m/s}} = \frac{1}{3 \times 10^{8}} \, \text{s} \]

### Step 3: Calculate the Acceleration in the y-direction
The acceleration \( a_y \) experienced by the charged particles due to the electric field can be calculated using:
\[ a_y = \frac{qE}{m} \]
Using the charge-to-mass ratio:
\[ a_y = \frac{q}{m} E = \left(2 \times 10^{11} \, \text{C/kg}\right) \left(1.8 \times 10^{3} \, \text{V/m}\right) \]
Calculating this gives:
\[ a_y = 3.6 \times 10^{14} \, \text{m/s}^2 \]

### Step 4: Calculate the Deflection in the y-direction
Using the kinematic equation for displacement in the y-direction:
\[ y = u_i t + \frac{1}{2} a_y t^2 \]
Since the initial velocity in the y-direction \( u_i = 0 \):
\[ y = \frac{1}{2} a_y t^2 \]
Substituting the values of \( a_y \) and \( t \):
\[ y = \frac{1}{2} \left(3.6 \times 10^{14} \, \text{m/s}^2\right) \left(\frac{1}{3 \times 10^{8}} \, \text{s}\right)^2 \]
Calculating \( t^2 \):
\[ t^2 = \left(\frac{1}{3 \times 10^{8}}\right)^2 = \frac{1}{9 \times 10^{16}} \, \text{s}^2 \]
Now substituting this back into the equation for \( y \):
\[ y = \frac{1}{2} \left(3.6 \times 10^{14}\right) \left(\frac{1}{9 \times 10^{16}}\right) \]
\[ y = \frac{3.6 \times 10^{14}}{18 \times 10^{16}} = 2 \times 10^{-2} \, \text{m} = 0.02 \, \text{m} = 20 \, \text{mm} \]

### Final Answer
The deflection of the charged particles in the y-direction is **20 mm**.