A particle of mass m and charge q moves at high speed along the x-axis. It is initially near `x = -oo` and it ends up near `x=+oo`. A second charge is fixed at the point x = 0, y=-d. As the moving charge passes the stationary charge, its x component of velocity does not change appreciably, but it acquires a small velocity in y-direction. Determine the angle through which the moving charge is deflected

Options

A. `theta=tan^(-1)((qQ)/(2piepsilon_0dmv^2))`
B. `theta=sin^(-1)((qQ)/(2piepsilon_0dmv^2))`
C. `theta=tan^(-1)((qQ)/(4piepsilon_0dmv^2))`
D. `theta=sin^(-1)((qQ)/(4piepsilon_0dmv^2))`

Answer

To solve the problem of the deflection of a charged particle moving in the electric field of a stationary charge, we can follow these steps:

### Step 1: Understand the Forces Acting on the Particle
The moving charge \( q \) experiences a force due to the electric field created by the stationary charge \( Q \) located at \( (0, -d) \). The electric field \( \vec{E} \) at the position of the moving charge can be calculated using Coulomb's law.

**Hint:** Remember that the electric field due to a point charge is given by \( \vec{E} = \frac{kQ}{r^2} \hat{r} \), where \( k \) is Coulomb's constant and \( r \) is the distance from the charge.

### Step 2: Calculate the Electric Field
The distance \( r \) from the stationary charge to the moving charge when it is at position \( (x, 0) \) is given by:
\[ r = \sqrt{x^2 + d^2} \]
The electric field \( \vec{E} \) at the position of the moving charge is:
\[ \vec{E} = \frac{kQ}{(x^2 + d^2)} \hat{r} \]
The direction of \( \hat{r} \) is from the stationary charge to the moving charge.

**Hint:** The components of the electric field can be broken down into \( E_x \) and \( E_y \).

### Step 3: Determine the Force on the Moving Charge
The force \( \vec{F} \) acting on the moving charge \( q \) due to the electric field is given by:
\[ \vec{F} = q \vec{E} \]
This force will have both x and y components:
- \( F_x = q E_x \)
- \( F_y = q E_y \)

**Hint:** Use the components of the electric field to find \( E_x \) and \( E_y \).

### Step 4: Analyze the Motion in the y-direction
Since the x-component of the velocity does not change appreciably, we can focus on the y-direction. The acceleration \( a_y \) of the charge in the y-direction can be found using Newton's second law:
\[ F_y = m a_y \implies a_y = \frac{F_y}{m} \]

**Hint:** Remember that acceleration is the change in velocity over time.

### Step 5: Calculate the Deflection Angle
The deflection angle \( \theta \) can be found using the relationship between the y-component of velocity \( v_y \) acquired and the x-component of velocity \( v_x \):
\[ \tan(\theta) = \frac{v_y}{v_x} \]
Since \( v_x \) is approximately constant, we can express \( v_y \) in terms of the acceleration and the time \( t \) it takes for the charge to pass the stationary charge.

**Hint:** Use the kinematic equation \( v_y = a_y t \) to relate the velocities.

### Step 6: Final Expression for the Angle
Substituting the expressions for \( v_y \) and \( v_x \) into the tangent equation gives:
\[ \theta \approx \tan^{-1}\left(\frac{a_y t}{v_x}\right) \]
This will provide the angle through which the moving charge is deflected.

**Hint:** For small angles, \( \tan(\theta) \approx \theta \) in radians.

### Summary
The angle of deflection \( \theta \) can be determined by analyzing the forces acting on the charged particle as it moves through the electric field of the stationary charge, calculating the resulting acceleration, and relating the velocities in the x and y directions.