Question

An electron falls from rest through a vertical distance h in a uniform and vertically upwards 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 it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall proton, is.

Answer 1

Hint: The electron that falls from rest, so its initial velocity will be zero. The electric field is evident in this region. Hence the electron will be experiencing electric force. Any charged particle in a region of the electric field experiences an electric force that will not cause the charge to be in motion.


Complete step by step solution:
As we know that electron of charge $e$ experiences an electric force due to the presence of the electric field $E$so, the force that acts on an electron is
In the field $E$ is upward, so the negatively charged $e$experiences a downward force.
The force experienced by electron of mass \[m\] is given by
\[\;F{\text{ }} = {\text{ }}e\,E\]
But,
\[F = m\,a\]
\[ \Rightarrow e\,E = m\,a\]
Therefore $a = \dfrac{{e\,E}}{m}$
here \[m\] is the mass of the electron.
Because electron starts to move it is in rest, the initial velocity \[u = 0\]
Now, use
\[S = ut + \dfrac{1}{2}{\text{ }}a{t^2}\]
\[ \Rightarrow h = 0 + \dfrac{1}{2}\left( {\dfrac{{eE}}{m}} \right){t^2}\]

The time required by the electron to fall over a distance \[h\] is given as:
\[t{\text{ }} = {\text{ }}\sqrt {\dfrac{{2hm}}{{eE}}} \]

Similarly in the case of the proton,
\[t'{\text{ }} = {\text{ }}\sqrt {\dfrac{{2hM}}{{pE}}} \]
Here\[M\] is the mass of the proton
We also know, the magnitude of the charge on protons equals electrons.
So, \[e = p\]
\[\therefore t'{\text{ }} = {\text{ }}\sqrt {\dfrac{{2hM}}{{eE}}} \]

here it is clear that,
\[\dfrac{t}{{t'{\text{ }}}}{\text{ }} = {\text{ }}\sqrt {\dfrac{m}{M}} \]
But mass of proton >> mass of the electron
∴\[M{\text{ }} > > {\text{ }}m\]
∴\[\dfrac{t}{{t'{\text{ }}}}{\text{ }} < {\text{ }}1\]
\[t < t'{\text{ }}\]
Hence, the time taken by an electron to fall is smaller than the time taken by a proton.

Note:Thus, the heavier particle that is the proton takes a greater time to fall through the same distance. In the case of the free fall of an object under gravity, the time of fall of an object is independent of the mass of the body. The acceleration of electrons and protons is greater than the acceleration due to gravity. Thus, the consequence of acceleration due to gravity can be ignored in this case.