
An electron of mass ${M_e}$initially at rest moves through a certain distance in a uniform electric field in time ${t_1}$. A proton of mass ${M_p}$also initially at rest takes time ${t_2}$to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of $\dfrac{{{t_2}}}{{{t_1}}}$ is nearly equal to
(A) $1$
(B) $\sqrt {\dfrac{{{M_p}}}{{{M_e}}}} $
(C) $\sqrt {\dfrac{{{M_e}}}{{{M_p}}}} $
(D) 1836
Answer
564.3k+ views
Hint
It is given that electrons and protons move in electric fields, and we know that a particle moving in an electric field experiences a force. We can use this to calculate the ratio.
$\Rightarrow F = qE$
$\Rightarrow S = \dfrac{1}{2}a{t^2}$
Complete step by step answer
When electron and proton will move in the electric field, then they will experience a force F which is given by
$\Rightarrow F = qE$
And from Newton’s laws of motion $F = ma$, where m is the mass of the body and a is the acceleration
Let ${F_e},{a_e},q$be the force, acceleration and charge of electron
So, ${M_e}{a_e} = qE \Rightarrow {a_e} = \dfrac{{qE}}{{{M_e}}}$ (the mass of electron was given)
Similarly Let ${F_p},{a_p},q$be the force, acceleration and charge of proton (the magnitude of charge on electron and proton are same)
So, ${M_p}{a_p} = qE \Rightarrow {a_p} = \dfrac{{qE}}{{{M_p}}}$
It is given that both electron and proton travel same distance let’s say S
By the equation of motion,
Distance travelled S by a body with acceleration a within time t is given by
$\Rightarrow S = \dfrac{1}{2}a{t^2}$
On substituting the given values, the equation becomes,
$\Rightarrow S = \dfrac{1}{2}\dfrac{{Eq}}{{{M_e}}}{t_1}^2 = \dfrac{1}{2}\dfrac{{Eq}}{{{M_p}}}{t_2}^2$
$\Rightarrow \dfrac{{{t_1}}}{{{t_2}}} = \sqrt {\dfrac{{{M_p}}}{{{M_e}}}} $
Hence the correct option is (B).
Additional Information
Electric field intensity at any point is the electrostatic force experienced by a unit positive charge in magnitude and direction. The SI unit of electric field intensity is N/C.
Electric field follows superposition principle, which states that in a system of charge, the field intensity can be calculated by the vector sum of the individual point charge field intensities.
Note
The direction of force on an electron in an electric field will be opposite to the direction of an electric field while in the case of protons, the direction of force will be in the same direction as that of the electric field.
It is given that electrons and protons move in electric fields, and we know that a particle moving in an electric field experiences a force. We can use this to calculate the ratio.
$\Rightarrow F = qE$
$\Rightarrow S = \dfrac{1}{2}a{t^2}$
Complete step by step answer
When electron and proton will move in the electric field, then they will experience a force F which is given by
$\Rightarrow F = qE$
And from Newton’s laws of motion $F = ma$, where m is the mass of the body and a is the acceleration
Let ${F_e},{a_e},q$be the force, acceleration and charge of electron
So, ${M_e}{a_e} = qE \Rightarrow {a_e} = \dfrac{{qE}}{{{M_e}}}$ (the mass of electron was given)
Similarly Let ${F_p},{a_p},q$be the force, acceleration and charge of proton (the magnitude of charge on electron and proton are same)
So, ${M_p}{a_p} = qE \Rightarrow {a_p} = \dfrac{{qE}}{{{M_p}}}$
It is given that both electron and proton travel same distance let’s say S
By the equation of motion,
Distance travelled S by a body with acceleration a within time t is given by
$\Rightarrow S = \dfrac{1}{2}a{t^2}$
On substituting the given values, the equation becomes,
$\Rightarrow S = \dfrac{1}{2}\dfrac{{Eq}}{{{M_e}}}{t_1}^2 = \dfrac{1}{2}\dfrac{{Eq}}{{{M_p}}}{t_2}^2$
$\Rightarrow \dfrac{{{t_1}}}{{{t_2}}} = \sqrt {\dfrac{{{M_p}}}{{{M_e}}}} $
Hence the correct option is (B).
Additional Information
Electric field intensity at any point is the electrostatic force experienced by a unit positive charge in magnitude and direction. The SI unit of electric field intensity is N/C.
Electric field follows superposition principle, which states that in a system of charge, the field intensity can be calculated by the vector sum of the individual point charge field intensities.
Note
The direction of force on an electron in an electric field will be opposite to the direction of an electric field while in the case of protons, the direction of force will be in the same direction as that of the electric field.
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