Question

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of the electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in through the same vertical distance h. The time fall of the electron, in comparison to the time fall of the proton is
1- smaller
2- five times greater
3- 10 times greater
4- equal

Answer 1

Hint: The electron falls from rest, so, its initial velocity is zero. The electric field is present in the region, so the electron will experience electric force. Any charged particle in a region of electric field experiences electric force which does not cause the charge to be in motion.

Complete step by step answer:
The vertical distance to be covered is h. The electric field magnitude is E and it is directed upwards. Since the electron is negatively charged, it will experience force downwards.
Electric force= \[qE\]
\[F=eE\]
Using newton’s second law, \[{{m}_{e}}a=eE\]
\[a=\dfrac{eE}{{{m}_{e}}}\]. Now using second equation of motion we get,
$
h=ut+\dfrac{a{{t}^{2}}}{2} \\
\implies h=0+\dfrac{a{{t}^{2}}}{2} \\
\implies h=\dfrac{a{{t}^{2}}}{2} \\
\implies h=\dfrac{\dfrac{eE}{{{m}_{e}}}\times {{t}^{2}}}{2} \\
\therefore h=\dfrac{eE{{t}^{2}}}{2{{m}_{e}}} \\
$
Finding time from this, \[t=\sqrt{\dfrac{2{{m}_{e}}h}{eE}}\]----------(1)
Now the proton takes the place of a neutron and the mass of the proton is greater than the mass of the electron but the magnitude of charge is the same. From eq (1) it is clear that \[t\alpha \sqrt{m}\], so the time taken by the proton will be greater than taken by the electron.

So, the correct answer is “Option 1”.

Note:
Both electron and proton have the same magnitude of charge but of opposite polarity. The charge on protons is positive and the charge on electrons is negative. Both the proton and electron experience electric force. Also, in this problem field was from down to top, the force experienced by the electron was downwards and that by the proton was upwards.