give answer in 2 step with explanation at the end of each step and final answer at the end:Part of the deflection system displayed in Fig. 1 employs vertically oriented electric fields
E = Ek with va

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:Part of the deflection system displayed in Fig. 1 employs vertically oriented electric fields
E = Ek with variable E to manipulate the motion of charged particles. Ar time 1 = 0,
when the applied field is some unknown E = E , a positively charged particle of charge ¢
and mass mx - axis in the figure, . Then, at a given time t = on the applied field is suddenly switched ro
another
unknown E = E, . Afterwards, ar another given time r = ot, this particle is observed
at point B, of the same vertical position as point A, but moving now ar angle « away from
the horizontal. Based on these observations, find the electric fields E and E, , including
their signs, that were used ro effect the motion of the particle? What speed / does the
particle have at point B? (Gravity can be neglected.)
[|
1. Part of the deflection system displayed in Fig. 1 employs vertically oriented electric fields
E = Ek with variable E' to manipulate the motion of charged particles. At time ¢ = 0,
when the applied field is some unknown E = E\, a positively charged particle of charge ¢
and mass m is observed at poi

VIPSolutions · Accepted Answer

## Step 1: Analyze Motion During Each Field

### **Explanation:**
- At $ t= $, the particle is at point A, moving horizontally (along the x-axis) with speed $ v_i $. The electric field $ E_1 $ is applied vertically.
- The force on the particle: $ F = qE_1 $, causing acceleration $ a_1 = \frac{qE_1}{m} $ in the z-direction.
- For time $ t_1 $, the particle’s velocity in $ z $: $ v_{z1} = a_1 t_1 = \frac{qE_1}{m} t_1 $.
- In x-direction, velocity remains $ v_i $.
- At $ t_1 $, the field switches to $ E_2 $ for an additional time $ t_2 $.
- The new acceleration: $ a_2 = \frac{qE_2}{m} $.
- Additional z-velocity gained: $ v_{z2} = a_2 t_2 = \frac{qE_2}{m} t_2 $.
- **Total z-velocity at B:** $ v_{zf} = v_{z1} + v_{z2} = \frac{qE_1}{m} t_1 + \frac{qE_2}{m} t_2 $.
- **Final velocity:** $ v_f = \sqrt{v_i^2 + v_{zf}^2} $.

**At the end of Step 1:**
- The particle’s final velocity components at B are:
  - $ v_x = v_i $
  - $ v_z = \frac{qE_1}{m} t_1 + \frac{qE_2}{m} t_2 $

---

## Step 2: Use Geometry and Final Angle

### **Explanation:**
- At point B, the particle moves at an angle $ \alpha $ to the horizontal: 
  $$
  \tan \alpha = \frac{v_{zf}}{v_i}
  $$
- Substitute $ v_{zf} $:
  $$
  \tan \alpha = \frac{\frac{qE_1}{m} t_1 + \frac{qE_2}{m} t_2}{v_i}
  $$
- Also, the particle ends at the same vertical position, meaning net vertical displacement = .
- Vertical displacement during $ t_1 $: 
  $$
  \Delta z_1 = \frac{1}{2} a_1 t_1^2 = \frac{1}{2} \frac{qE_1}{m} t_1^2
  $$
- During $ t_2 $:
  $$
  \Delta z_2 = v_{z1} t_2 + \frac{1}{2} a_2 t_2^2 = \frac{qE_1}{m} t_1 t_2 + \frac{1}{2} \frac{qE_2}{m} t_2^2
  $$
- Net vertical displacement:
  $$
  \Delta z = \Delta z_1 + \Delta z_2 = 
  $$
  $$
  \frac{1}{2} \frac{qE_1}{m} t_1^2 + \frac{qE_1}{m} t_1 t_2 + \frac{1}{2} \frac{qE_2}{m} t_2^2 = 
  $$
  $$
  E_1 ( \frac{1}{2} t_1^2 + t_1 t_2 ) + \frac{1}{2} E_2 t_2^2 = 
  $$

**At the end of Step 2:**
- You have two equations:
  1. $ \tan \alpha = \frac{\frac{qE_1}{m} t_1 + \frac{qE_2}{m} t_2}{v_i} $
  2. $ E_1 ( \frac{1}{2} t_1^2 + t_1 t_2 ) + \frac{1}{2} E_2 t_2^2 =  $

---

## **Final Answers**

1. **Electric fields $ E_1 $ and $ E_2 $:**
   $$
   \boxed{
   \begin{align*}
   \tan \alpha &= \frac{\frac{qE_1}{m} t_1 + \frac{qE_2}{m} t_2}{v_i} \
   E_1 ( \frac{1}{2} t_1^2 + t_1 t_2 ) + \frac{1}{2} E_2 t_2^2 &= 
   \end{align*}
   }
   $$
   Solve these equations simultaneously for $ E_1 $ and $ E_2 $.

2. **Final speed $ v_f $ at point B:**
   $$
   \boxed{
   v_f = \sqrt{v_i^2 + \left( \frac{qE_1}{m} t_1 + \frac{qE_2}{m} t_2 \right)^2 }
   }
   $$

**Signs:**  
The sign of $ E_1 $ and $ E_2 $ will depend on the direction needed to achieve the required final angle and zero net displacement. One will be positive, and the other negative if the fields must reverse direction to return the particle to the original vertical level.

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