An electron falls from rest through a vertical distance h in a uniform and vertically upward 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 flal of the 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 an electron and a proton falling through the same vertical distance \( h \) in an electric field \( E \). Let's break down the solution step by step.

### Step 1: Understand the Forces Acting on the Electron and Proton
- The force acting on the electron (\( F_e \)) in the upward electric field is given by:
\[ F_e = -eE \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \, C \)).

- The force acting on the proton (\( F_p \)) when the electric field is reversed (downward) is:
\[ F_p = eE \]
where \( e \) is the charge of the proton (also approximately \( 1.6 \times 10^{-19} \, C \)).

### Step 2: Calculate the Acceleration of the Electron and Proton
- The acceleration of the electron (\( a_e \)) 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 (approximately \( 9.1 \times 10^{-31} \, kg \)).

- The acceleration of the proton (\( a_p \)) is:
\[ a_p = \frac{F_p}{m_p} = \frac{eE}{m_p} \]
where \( m_p \) is the mass of the proton (approximately \( 1.67 \times 10^{-27} \, kg \)).

### Step 3: Use the Equation of Motion
- The distance fallen by both the electron and proton is \( h \). Using the equation of motion \( s = ut + \frac{1}{2} a t^2 \) and noting that the initial velocity \( u = 0 \):
\[ h = \frac{1}{2} a_e t_e^2 \quad \text{(for the electron)} \]
\[ h = \frac{1}{2} a_p t_p^2 \quad \text{(for the proton)} \]

### Step 4: Relate the Times of Fall
- From the equations above, we can express the time of fall for the electron and proton:
\[ t_e^2 = \frac{2h}{a_e} \quad \text{and} \quad t_p^2 = \frac{2h}{a_p} \]
- Taking the ratio of the squares of the times:
\[ \frac{t_e^2}{t_p^2} = \frac{a_p}{a_e} \]

### Step 5: Substitute the Accelerations
- Substitute the expressions for \( a_e \) and \( a_p \):
\[ \frac{t_e^2}{t_p^2} = \frac{\frac{eE}{m_p}}{\frac{-eE}{m_e}} = \frac{m_e}{m_p} \]
- This implies:
\[ \frac{t_e}{t_p} = \sqrt{\frac{m_e}{m_p}} \]

### Step 6: Determine the Relationship Between the Times
- Since \( m_e \) (mass of the electron) is much smaller than \( m_p \) (mass of the proton):
\[ \frac{t_e}{t_p} < 1 \]
- This means that the time of fall of the electron \( t_e \) is less than the time of fall of the proton \( t_p \).

### Conclusion
Thus, the time of fall of the electron, in comparison to the time of fall of the proton, is shorter.