The height of a tower is h. The acceleration due to gravity is g. Everywhere in the surroundings of the twowr there is a uniform electric field of intensity E in the horizontal direction away from the tower. A particle of mass m and carryign a charge q is dropped from the top of tower. The distance of the partile when it reaches the ground from the foot of the tower (neglect the effect of air on the motion of the particle )

Options

A. `(qEh)/(mg)`
B. `(qEhg)/m`
C. `(mg)/(qEH)`
D. `m/(qEgh)`

Explanation

To solve the problem step by step, we will analyze the motion of the charged particle dropped from the tower in both vertical and horizontal directions.

### Step 1: Understand the Motion
The particle is dropped from a height \( h \) with an initial velocity of zero. It experiences two forces: the gravitational force acting downward and the electric force acting horizontally due to the uniform electric field \( E \).

### Step 2: Identify Forces Acting on the Particle
1. **Gravitational Force**: The force due to gravity is given by:
\[ F_g = mg \]
where \( m \) is the mass of the particle and \( g \) is the acceleration due to gravity.

2. **Electric Force**: The force acting on the particle due to the electric field is:
\[ F_e = qE \]
where \( q \) is the charge of the particle and \( E \) is the electric field intensity.

### Step 3: Calculate Acceleration in the Horizontal Direction
Using Newton's second law, the acceleration \( a_x \) in the horizontal direction can be calculated as:
\[ a_x = \frac{F_e}{m} = \frac{qE}{m} \]

### Step 4: Calculate Time of Flight
To find the time \( t \) it takes for the particle to reach the ground, we can use the equation of motion in the vertical direction:
\[ S_y = u_y t + \frac{1}{2} a_y t^2 \]
Here, \( S_y = -h \) (downward), \( u_y = 0 \) (initial vertical velocity), and \( a_y = -g \) (acceleration due to gravity). Thus, we have:
\[ -h = 0 \cdot t + \frac{1}{2} (-g) t^2 \]
This simplifies to:
\[ h = \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t^2 = \frac{2h}{g} \quad \Rightarrow \quad t = \sqrt{\frac{2h}{g}} \]

### Step 5: Calculate Horizontal Displacement
Now, we can calculate the horizontal displacement \( x \) using the formula:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
Since the initial horizontal velocity \( u_x = 0 \), this simplifies to:
\[ x = \frac{1}{2} a_x t^2 \]
Substituting \( a_x = \frac{qE}{m} \) and \( t = \sqrt{\frac{2h}{g}} \):
\[ x = \frac{1}{2} \left(\frac{qE}{m}\right) \left(\sqrt{\frac{2h}{g}}\right)^2 \]
This simplifies to:
\[ x = \frac{1}{2} \left(\frac{qE}{m}\right) \left(\frac{2h}{g}\right) \]
Thus, we have:
\[ x = \frac{qEh}{mg} \]

### Final Answer
The distance of the particle from the foot of the tower when it reaches the ground is:
\[ x = \frac{qEh}{mg} \]

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