An electron falls from rest through a vertical distance h in a uniform and vertically upwards directed electric field E. The direction of electric field is now reversed , keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall proton is

Options

A. 10 times greater
B. 5 times greater
C. Smaller
D. Equal

Explanation

To solve the problem, we need to analyze the motion of both the electron and the proton under the influence of an electric field. Let's break it down step by step.

### Step 1: Understanding the Forces
1. **Electron in Upward Electric Field**: When the electron falls through a distance \( h \) in an upward electric field \( E \), the force acting on it is due to the electric field, which is given by:
\[ F_e = -eE \]
where \( e \) is the charge of the electron (positive value, but the force is negative due to the negative charge of the electron).

2. **Proton in Downward Electric Field**: When the electric field is reversed, the proton falls through the same distance \( h \). The force acting on the proton is:
\[ F_p = +eE \]
where \( e \) is the charge of the proton.

### Step 2: Calculating Acceleration
1. **Acceleration of Electron**: The acceleration \( a_e \) of the electron can be calculated using Newton's second law:
\[ a_e = \frac{F_e}{m_e} = \frac{-eE}{m_e} \]
where \( m_e \) is the mass of the electron.

2. **Acceleration of Proton**: Similarly, the acceleration \( a_p \) of the proton is:
\[ a_p = \frac{F_p}{m_p} = \frac{eE}{m_p} \]
where \( m_p \) is the mass of the proton.

### Step 3: Using the Equation of Motion
Since both particles start from rest, we can use the second equation of motion:
\[ h = \frac{1}{2} a t^2 \]
to find the time taken for each particle to fall the distance \( h \).

1. **Time for Electron**:
\[ h = \frac{1}{2} \left(\frac{-eE}{m_e}\right) t_e^2 \implies t_e^2 = \frac{2h m_e}{eE} \implies t_e = \sqrt{\frac{2h m_e}{eE}} \]

2. **Time for Proton**:
\[ h = \frac{1}{2} \left(\frac{eE}{m_p}\right) t_p^2 \implies t_p^2 = \frac{2h m_p}{eE} \implies t_p = \sqrt{\frac{2h m_p}{eE}} \]

### Step 4: Comparing Times
Now, we can compare the times \( t_e \) and \( t_p \):
\[ \frac{t_e}{t_p} = \sqrt{\frac{m_e}{m_p}} \]
Since the mass of the electron \( m_e \) is much smaller than the mass of the proton \( m_p \) (approximately \( m_e \approx 9.11 \times 10^{-31} \, \text{kg} \) and \( m_p \approx 1.67 \times 10^{-27} \, \text{kg} \)), we conclude that:
\[ t_e < t_p \]

### Final Conclusion
The time of fall of the electron is less than the time of fall of the proton.