A charged bead is capable of sliding freely through a string held vertically in tension. An electric field is applied parallel to the string so that the bead stays at rest of the middle of the string. If the electric field is switched off momentarily and switched on

Options

A. the bead moves downwards and stops as soon as the field is switched on
B. the bead moved downwards when the field is switched off and moves upwards when the field is switched on
C. the bead moves downwards with constant acceleration till it reaches the bottom of the string
D. the bead moves downwards with constant velocity till it reaches the bottom of the string

Answer

To solve the problem, we need to analyze the behavior of the charged bead when the electric field is switched off and then switched back on. Here’s a step-by-step solution:

### Step 1: Understand the Initial Condition
The charged bead is at rest in the middle of the string due to the balance of forces. The electric force acting on the bead (due to the electric field) is equal to the gravitational force acting on it.

**Hint:** Remember that the forces acting on the bead are balanced when it is at rest.

### Step 2: Write the Force Balance Equation
When the bead is at rest, we can write the equation:
\[ E \cdot Q = mg \]
Where:
- \( E \) is the electric field strength,
- \( Q \) is the charge of the bead,
- \( m \) is the mass of the bead,
- \( g \) is the acceleration due to gravity.

**Hint:** This equation shows that the electric force equals the weight of the bead.

### Step 3: Analyze the Situation When the Electric Field is Switched Off
When the electric field is switched off momentarily, the only force acting on the bead is the gravitational force \( mg \). The bead will start to accelerate downwards due to gravity.

**Hint:** Think about how the bead will behave when only gravity is acting on it.

### Step 4: Calculate the Velocity of the Bead
Since the bead starts from rest, we can use the equation of motion to find its velocity after a time \( t \):
\[ v = u + gt \]
Where:
- \( u = 0 \) (initial velocity),
- \( v \) is the final velocity after time \( t \).

Thus, we have:
\[ v = gt \]

**Hint:** The bead accelerates downwards under gravity, so its velocity increases linearly with time.

### Step 5: Analyze the Situation When the Electric Field is Switched On Again
When the electric field is switched back on, the electric force will act on the bead again. However, at this moment, the bead has a downward velocity \( v = gt \). The electric force will now balance the gravitational force, but since the bead already has downward velocity, it will continue moving downwards.

**Hint:** Consider what happens to the forces acting on the bead when the electric field is reapplied.

### Step 6: Determine the Motion of the Bead
Since the electric force re-establishes the balance with the gravitational force, the net force on the bead becomes zero. Therefore, the bead will move downwards with a constant velocity \( gt \) until it reaches the bottom of the string.

**Hint:** Remember that constant velocity means no net acceleration.

### Final Conclusion
The correct answer is that the bead moves downwards with constant velocity until it reaches the bottom of the string.

### Answer:
**Option 4: The bead moves downwards with constant velocity till it reaches the bottom of the string.**