An electron falls through a small distance in a uniform electric field of magnitude `2xx10^(4)NC^(-1)`. The direction of the field reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be

Options

A. Same in both cases
B. More in the case of electron
C. More in the case of proton
D. Independent of charge

Explanation

To solve the problem, we need to analyze the motion of both the electron and the proton in a uniform electric field. Let's go through the steps systematically.

### Step 1: Understanding the Forces
When an electron is placed in a uniform electric field \( E \), it experiences a force given by:
\[ F = qE \]
where \( q \) is the charge of the electron. The charge of the electron is \( -1.6 \times 10^{-19} \, C \). The direction of the force on the electron is opposite to the direction of the electric field because the electron has a negative charge.

For a proton, which has a charge of \( +1.6 \times 10^{-19} \, C \), the force will be in the same direction as the electric field.

### Step 2: Calculating the Acceleration
Using Newton's second law, the acceleration \( a \) of the electron can be calculated as:
\[ a_e = \frac{F}{m_e} = \frac{qE}{m_e} \]
where \( m_e \) is the mass of the electron, approximately \( 9.11 \times 10^{-31} \, kg \).

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

### Step 3: Finding the Time of Fall
Assuming both particles start from rest and fall through a distance \( s \), we can use the equation of motion:
\[ s = \frac{1}{2} a t^2 \]
Rearranging this gives:
\[ t = \sqrt{\frac{2s}{a}} \]

For the electron:
\[ t_e = \sqrt{\frac{2s}{a_e}} = \sqrt{\frac{2s m_e}{qE}} \]

For the proton:
\[ t_p = \sqrt{\frac{2s}{a_p}} = \sqrt{\frac{2s m_p}{qE}} \]

### Step 4: Comparing the Times
Now, we can compare \( t_e \) and \( t_p \):
\[ t_e = \sqrt{\frac{2s m_e}{qE}} \quad \text{and} \quad t_p = \sqrt{\frac{2s m_p}{qE}} \]

Since \( m_e \) (mass of the electron) is much smaller than \( m_p \) (mass of the proton), it follows that:
\[ t_e < t_p \]

### Conclusion
Thus, the time taken for the electron to fall through the distance \( s \) is less than the time taken for the proton to fall through the same distance in the electric field.

### Final Answer
The time of fall for the electron is less than that for the proton.

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