Question

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 as compared to the time of fall of the proton is

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

Answer 1

The correct option is B smaller

For electron:-

$m_{e}g+F_e=m_{e}a$

$a=\frac{m_{e}g+eE}{m_e}$

$a_e=g+\frac{e}{m_e}E$

Now, $h=\frac{1}{2}{a}_e{t}^2$

$t_e=\sqrt{\frac{2h}{g+\frac{eE}{m_e}}}$

For proton:
$m_{p}g+F_e=m_{p}a$

$a_p=g+\frac{eE}{m_p}$
Now, $h=\frac{1}{2}\times a_p{t}^2$
$\Rightarrow t_p=\sqrt{\frac{2h}{g+\frac{eE}{m_p}}}$
$\because m_p>m_e$
$t_e<t_p$