A uniform rod of length `l` and mass m is given a charge Q and is suspended vertically by means of a hinge at the top end. A horizontal electric field E is switched on, in the direction in which the rod can sway freely. Find the angle made by the rod with the vertical in equilibrium.

Answer

To solve the problem, we need to analyze the forces acting on the uniformly charged rod when it is subjected to a horizontal electric field. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Rod
The rod experiences two main forces:
1. **Weight (W)**: The weight of the rod acts downward and is given by \( W = mg \), where \( m \) is the mass of the rod and \( g \) is the acceleration due to gravity.
2. **Electric Force (F)**: The electric force acting on the rod due to the electric field \( E \) is given by \( F = QE \), where \( Q \) is the charge on the rod.

### Step 2: Set Up the Equilibrium Condition
In equilibrium, the rod will make an angle \( \theta \) with the vertical. The forces acting on the rod can be resolved into components. The weight acts vertically downward, while the electric force acts horizontally.

### Step 3: Analyze the Forces
At equilibrium, the tangent of the angle \( \theta \) can be expressed as the ratio of the electric force to the weight of the rod:
\[ \tan(\theta) = \frac{\text{Electric Force}}{\text{Weight}} = \frac{F}{W} \]
Substituting the expressions for the electric force and weight, we have:
\[ \tan(\theta) = \frac{QE}{mg} \]

### Step 4: Solve for the Angle \( \theta \)
To find the angle \( \theta \), we take the arctangent of both sides:
\[ \theta = \tan^{-1}\left(\frac{QE}{mg}\right) \]

### Final Answer
Thus, the angle made by the rod with the vertical in equilibrium is:
\[ \theta = \tan^{-1}\left(\frac{QE}{mg}\right) \]