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Question: An electron falls through a distance of 1.5cm in a uniform electric field of magnitude \(2 \times {1...

An electron falls through a distance of 1.5cm in a uniform electric field of magnitude 2×104N/C2 \times {10^4}N/C. Now, the direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. Compute the time of fall in each case, neglecting gravity.

Explanation

Solution

Hint: First, we will use our knowledge of the values of charge and mass of proton and electron in order to find the force and then the acceleration of both proton and electron respectively. Then we will use the second equation of motion to find out the time taken in both the cases. Refer to the solution below.

Formula used: F⃗=qE⃗\vec F = q\vec E, S=ut+12at2S = ut + \dfrac{1}{2}a{t^2}.

Complete step-by-step solution -

According to the question, the distance covered by the electron while falling is-
⇒S=1.5cm  ⇒S=1.5×10−2m  \Rightarrow S = 1.5cm \\\ \\\ \Rightarrow S = 1.5 \times {10^{ - 2}}m \\\
As we know that-
The mass of an electron is me=9.10×10−31kg{m_e} = 9.10 \times {10^{ - 31}}kg
The mass of a proton is mp=1.672×10−27kg{m_p} = 1.672 \times {10^{ - 27}}kg
The formula of force in terms of charge is F⃗=qE⃗\vec F = q\vec E
The magnitude of the electric field as given in the question is-
⇒E⃗=2×104N/C\Rightarrow \vec E = 2 \times {10^4}N/C
The charge on an electron is qe=−1.6×10−19{q_e} = - 1.6 \times {10^{ - 19}}
The change on a proton is qp=1.6×10−19{q_p} = 1.6 \times {10^{ - 19}}
Putting these values in the formula of force, we get-
Force on proton-
⇒F⃗=1.6×10−19×2×104  ⇒F⃗=3.2×10−15N  \Rightarrow \vec F = 1.6 \times {10^{ - 19}} \times 2 \times {10^4} \\\ \\\ \Rightarrow \vec F = 3.2 \times {10^{ - 15}}N \\\
Force on electron-
⇒F⃗=−1.6×10−19×2×104  ⇒F⃗=−3.2×10−15N  \Rightarrow \vec F = - 1.6 \times {10^{ - 19}} \times 2 \times {10^4} \\\ \\\ \Rightarrow \vec F = - 3.2 \times {10^{ - 15}}N \\\
According to Newton’s second law, F⃗=ma⃗\vec F = m\vec a.
Value of acceleration for proton-
⇒ap=F⃗mp  ⇒ap=1.913×1012m/s2  \Rightarrow {a_p} = \dfrac{{\vec F}}{{{m_p}}} \\\ \\\ \Rightarrow {a_p} = 1.913 \times {10^{12}}m/{s^2} \\\
Value of acceleration for electron-
⇒ae=F⃗me  ⇒ae=−0.35×1016m/s2  \Rightarrow {a_e} = \dfrac{{\vec F}}{{{m_e}}} \\\ \\\ \Rightarrow {a_e} = - 0.35 \times {10^{16}}m/{s^2} \\\
The second equation of motion is S=ut+12at2S = ut + \dfrac{1}{2}a{t^2}.
According to the question, the direction of the field was reversed. So, we will ignore the sign convention for acceleration.
Putting the values in the second equation of motion for proton, we get-
⇒1.5×10−2=12×1.913×1012×tp2  ⇒1.568×10−14=tp2  ⇒tp=1.252×10−7sec⁡  \Rightarrow 1.5 \times {10^{ - 2}} = \dfrac{1}{2} \times 1.913 \times {10^{12}} \times {t_p}^2 \\\ \\\ \Rightarrow 1.568 \times {10^{ - 14}} = {t_p}^2 \\\ \\\ \Rightarrow {t_p} = 1.252 \times {10^{ - 7}}\sec \\\
Putting the values in the second equation of motion for electron, we get-
⇒1.5×10−2=12×3.5×1015×te2  ⇒0.857×10−17=te2  ⇒te=2.927×10−8sec⁡  \Rightarrow 1.5 \times {10^{ - 2}} = \dfrac{1}{2} \times 3.5 \times {10^{15}} \times {t_e}^2 \\\ \\\ \Rightarrow 0.857 \times {10^{ - 17}} = {t_e}^2 \\\ \\\ \Rightarrow {t_e} = 2.927 \times {10^{ - 8}}\sec \\\

Note: Ignore the sign convention whenever the direction of the field is reversed. You must remember the exact values of the masses of electron and proton as well as their charges. The charge of an electron will always be negative.