Question

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of electrio 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 of the proton is

• A. equal
• B. smaller
• C. 10 times greater
• D. 5 times greator

Answer 1

The Correct Option is B

To solve this problem, let's analyze the motion of both the electron and the proton when they fall through a vertical distance $h$ in a given electric field $E$.

1. Initially, consider the electron falling in an electric field $E$ directed upwards.

The force acting on the electron due to the electric field is given by:

F_e = eE

where e is the elementary charge.

2. According to Newton's second law, the acceleration a_e of the electron is:

a_e = \frac{F_e}{m_e} = \frac{eE}{m_e}

Here, m_e is the mass of the electron.

3. When the electron falls through height h, using the kinematic equation h = \frac{1}{2} a_e t_e^2 (since the electron starts from rest), we can solve for the time t_e:

t_e = \sqrt{\frac{2h}{a_e}} = \sqrt{\frac{2h \times m_e}{eE}}

4. Now, consider the proton in a reversed electric field with the same magnitude. The force on the proton is:

F_p = eE

where e is the same elementary charge (but the displacement does not affect calculation as magnitude is considered only).

5. The acceleration a_p of the proton is:

a_p = \frac{F_p}{m_p} = \frac{eE}{m_p}

Here, m_p is the mass of the proton.

6. Using the same kinematic equation for the proton, h = \frac{1}{2} a_p t_p^2, we solve for the time t_p:

t_p = \sqrt{\frac{2h}{a_p}} = \sqrt{\frac{2h \times m_p}{eE}}

7. Comparing the times t_e and t_p, we get:

\frac{t_e}{t_p} = \sqrt{\frac{m_e}{m_p}}

Since m_e \ll m_p, we have t_e \ll t_p. Therefore, the time of fall for the electron is smaller compared to the proton.

Thus, the correct answer is smaller.