To solve the problem, we need to analyze the forces acting on the charged particle and find the velocity required for it to continue moving horizontally without any vertical acceleration.
### Step 1: Identify the Forces Acting on the Particle
The forces acting on the particle are:
1. **Weight (W)**: Acts downward, given by \( W = mg \).
2. **Electric Force (F_e)**: Acts upward due to the electric field, given by \( F_e = qE \).
3. **Magnetic Force (F_b)**: Acts perpendicular to both the velocity and the magnetic field, given by \( F_b = qvB \).
### Step 2: Calculate the Weight of the Particle
Given:
- Mass \( m = 2 \times 10^{-5} \, \text{kg} \)
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)
The weight of the particle is:
\[
W = mg = (2 \times 10^{-5} \, \text{kg})(9.8 \, \text{m/s}^2) = 1.96 \times 10^{-4} \, \text{N}
\]
### Step 3: Calculate the Electric Force
Given:
- Charge \( q = 10^{-6} \, \text{C} \)
- Electric field \( E = 200 \, \text{N/C} \)
The electric force is:
\[
F_e = qE = (10^{-6} \, \text{C})(200 \, \text{N/C}) = 2 \times 10^{-4} \, \text{N}
\]
### Step 4: Set Up the Equation for Horizontal Motion
For the particle to continue moving horizontally, the net vertical force must be zero. Therefore, the upward electric force must balance the downward weight and the magnetic force:
\[
F_e = W + F_b
\]
### Step 5: Express the Magnetic Force
The magnetic force can be expressed as:
\[
F_b = qvB
\]
where \( B = 2 \, \text{T} \).
### Step 6: Substitute Values into the Equation
Substituting the expressions for \( F_e \) and \( F_b \) into the balance equation:
\[
2 \times 10^{-4} = 1.96 \times 10^{-4} + (10^{-6})(v)(2)
\]
### Step 7: Solve for Velocity \( v \)
Rearranging the equation gives:
\[
2 \times 10^{-4} - 1.96 \times 10^{-4} = 2 \times 10^{-6} v
\]
\[
0.04 \times 10^{-4} = 2 \times 10^{-6} v
\]
\[
0.04 = 2v
\]
\[
v = \frac{0.04}{2} = 0.02 \, \text{m/s} = 2 \, \text{m/s}
\]
### Final Answer
The velocity of the charged particle so that it continues to move horizontally is:
\[
\boxed{2 \, \text{m/s}}
\]
