There is a uniform electric field of strength `10^(3)V//m` along `y`-axis. A body of mass `1g` and charge `10^(-6)C` is projected into the field from origin along the positive `x`-axis with a velocity `10 m//s`. Its speed in `m//s` after `10 s` is (Neglect gravitation)

Options

A. 10
B. `5sqrt(2)`
C. `10sqrt(2)`
D. 20

Answer

To solve the problem step by step, we need to analyze the motion of the charged body in the electric field. Here’s how we can approach it:

### Step 1: Understand the Problem
We have a uniform electric field of strength \( E = 10^3 \, \text{V/m} \) directed along the y-axis. A body with mass \( m = 1 \, \text{g} = 10^{-3} \, \text{kg} \) and charge \( q = 10^{-6} \, \text{C} \) is projected from the origin along the positive x-axis with an initial velocity \( u_x = 10 \, \text{m/s} \).

### Step 2: Calculate the Force Acting on the Body
The force \( F \) acting on the body due to the electric field is given by:
\[ F = qE \]
Substituting the values:
\[ F = (10^{-6} \, \text{C})(10^3 \, \text{V/m}) = 10^{-3} \, \text{N} \]

### Step 3: Calculate the Acceleration
Using Newton's second law, the acceleration \( a \) of the body can be calculated as:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{10^{-3} \, \text{N}}{10^{-3} \, \text{kg}} = 1 \, \text{m/s}^2 \]

### Step 4: Determine the Velocity in the Y-Direction After 10 Seconds
Since the body starts from rest in the y-direction, the initial velocity \( u_y = 0 \, \text{m/s} \). We can use the first equation of motion to find the final velocity \( v_y \) in the y-direction after \( t = 10 \, \text{s} \):
\[ v_y = u_y + at \]
Substituting the values:
\[ v_y = 0 + (1 \, \text{m/s}^2)(10 \, \text{s}) = 10 \, \text{m/s} \]

### Step 5: Calculate the Net Speed After 10 Seconds
The body has a constant velocity in the x-direction and a velocity in the y-direction after 10 seconds. The final velocities are:
- \( v_x = 10 \, \text{m/s} \) (in the x-direction)
- \( v_y = 10 \, \text{m/s} \) (in the y-direction)

The net speed \( v \) can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(10 \, \text{m/s})^2 + (10 \, \text{m/s})^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{m/s} \]

### Final Answer
The speed of the body after 10 seconds is \( 10\sqrt{2} \, \text{m/s} \).

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