Two point masses, m each carrying charges `-q` and `+q` are attached to the ends of a massless rigid non - conducting wire of length L. When this arrangement is placed in a uniform electric field, then it deflects through an angle `theta`. The minimum time needed by the rod to align itself along the field is

Options

A. `2pisqrt((mL)/(qE))`
B. `(pi)/(2)sqrt((mL)/(2qE))`
C. `pisqrt((2mL)/(qE))`
D. `2pi sqrt((3mL)/(qE))`

Answer

To solve the problem, we need to analyze the forces acting on the two point masses and how they lead to the torque that causes the deflection of the rod in the electric field. Here’s a step-by-step solution:

### Step 1: Understanding the Setup
We have two point masses, each of mass \( m \), attached to a massless rigid non-conducting wire of length \( L \). One mass carries a charge of \( -q \) and the other carries a charge of \( +q \). When placed in a uniform electric field \( E \), the arrangement will experience forces due to the electric field.

**Hint:** Visualize the system and identify the forces acting on each charge due to the electric field.

### Step 2: Calculate the Forces
The force acting on the charge \( +q \) is given by:
\[ F_+ = qE \]
The force acting on the charge \( -q \) is:
\[ F_- = -qE \]
Since the charges are equal in magnitude but opposite in sign, the magnitudes of the forces are the same:
\[ F = qE \]

**Hint:** Remember that the direction of the force on the positive charge is in the direction of the electric field, while the force on the negative charge is in the opposite direction.

### Step 3: Calculate the Torque
The torque \( \tau \) about the center of the rod (midpoint) due to these forces can be calculated. The distance from the center to each charge is \( \frac{L}{2} \). The torque due to each force is:
\[ \tau_+ = F_+ \cdot \frac{L}{2} \cdot \sin(\theta) = qE \cdot \frac{L}{2} \cdot \sin(\theta) \]
\[ \tau_- = F_- \cdot \frac{L}{2} \cdot \sin(\theta) = -(-qE) \cdot \frac{L}{2} \cdot \sin(\theta) = qE \cdot \frac{L}{2} \cdot \sin(\theta) \]
Thus, the total torque is:
\[ \tau = \tau_+ + \tau_- = qE \cdot \frac{L}{2} \cdot \sin(\theta) + qE \cdot \frac{L}{2} \cdot \sin(\theta) = qE L \sin(\theta) \]

**Hint:** Use the fact that torque is the product of force and the perpendicular distance from the pivot point.

### Step 4: Moment of Inertia
The moment of inertia \( I \) of the system about the center is:
\[ I = 2 \cdot m \left(\frac{L}{2}\right)^2 = 2 \cdot m \cdot \frac{L^2}{4} = \frac{mL^2}{2} \]

**Hint:** Recall that the moment of inertia for point masses is calculated using \( I = m r^2 \).

### Step 5: Relate Torque and Angular Acceleration
Using the relation \( \tau = I \alpha \), we can express the angular acceleration \( \alpha \):
\[ \alpha = \frac{\tau}{I} = \frac{qE L \sin(\theta)}{\frac{mL^2}{2}} = \frac{2qE \sin(\theta)}{mL} \]

**Hint:** This step connects the torque to the angular motion of the system.

### Step 6: Small Angle Approximation
For small angles, \( \sin(\theta) \approx \theta \) (in radians). Thus:
\[ \alpha \approx \frac{2qE \theta}{mL} \]

**Hint:** This approximation simplifies the calculations for small deflections.

### Step 7: Finding the Time Period
The angular frequency \( \omega \) can be derived from the equation of motion:
\[ \alpha = -\omega^2 \theta \]
Comparing both expressions gives:
\[ \omega^2 = \frac{2qE}{mL} \]
Thus, the time period \( T \) is:
\[ T = 2\pi \sqrt{\frac{I}{\tau}} = 2\pi \sqrt{\frac{mL^2/2}{qE}} = 2\pi \sqrt{\frac{mL}{2qE}} \]

**Hint:** The time period is the time taken for one complete oscillation.

### Step 8: Minimum Time to Align
To find the minimum time to align with the electric field, we consider that it takes \( \frac{T}{4} \) to align from an initial angle to the direction of the field:
\[ t_{\text{min}} = \frac{T}{4} = \frac{1}{4} \cdot 2\pi \sqrt{\frac{mL}{2qE}} = \frac{\pi}{2} \sqrt{\frac{mL}{2qE}} \]

**Hint:** This step uses the concept of oscillation to determine how quickly the system can align with the field.

### Final Solution
The minimum time needed for the rod to align itself along the electric field is:
\[ t_{\text{min}} = \frac{\pi}{2} \sqrt{\frac{mL}{2qE}} \]