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Question: In the uniform electric field of \(E = 1 \times {10^4}N{C^{ - 1}}\) , an electron is accelerated fro...

In the uniform electric field of E=1×104NC−1E = 1 \times {10^4}N{C^{ - 1}} , an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of 2×10−2m2 \times {10^{ - 2}}m m is nearly ____ ms−1m{s^{ - 1}}
(em of electron≃1.8×1011Ckg−1)\left( {\dfrac{e}{m}{\text{ of electron}} \simeq 1.8 \times {{10}^{11}}Ck{g^{ - 1}}} \right)

A. 1.6×106 B. 0.85×106 C. 0.425×106 D. 8.5×106  A.{\text{ }}1.6 \times {10^6} \\\ B.{\text{ }}0.85 \times {10^6} \\\ C.{\text{ }}0.425 \times {10^6} \\\ D.{\text{ }}8.5 \times {10^6} \\\
Explanation

Solution

Hint- In order to deal with this question we will use the third equation of motion but for applying it first we have to find the acceleration that will be found out through the formula which is mentioned in the solution. We will also use the concept of electrostatic in order to find the acceleration of the particle.

Formula used- a=Fm,F=eE,a=eEm,v2−u2=2asa = \dfrac{F}{m},F = eE,a = \dfrac{{eE}}{m},{v^2} - {u^2} = 2as

Completes step-by-step answer:

Given that: E=1×104NC−1E = 1 \times {10^4}N{C^{ - 1}} , distance covered by electron is 2×10−2m2 \times {10^{ - 2}}m and (em of electron≃1.8×1011Ckg−1)\left( {\dfrac{e}{m}{\text{ of electron}} \simeq 1.8 \times {{10}^{11}}Ck{g^{ - 1}}} \right)
As we know that acceleration of the electron is given as
a=Fma = \dfrac{F}{m}
Where F is the electrostatic force given by F=eEF = eE and m is the mass of electron
⇒a=eEm\Rightarrow a = \dfrac{{eE}}{m}
Substitute the given values in above formula we have
∵a=eEm ⇒a=(em)×E ⇒a=(1.8×1011Ckg−1)×1×104NC−1 [∵em of electron≃1.8×1011Ckg−1] ⇒a=1.8×1015m/s2  \because a = \dfrac{{eE}}{m} \\\ \Rightarrow a = \left( {\dfrac{e}{m}} \right) \times E \\\ \Rightarrow a = \left( {1.8 \times {{10}^{11}}Ck{g^{ - 1}}} \right) \times 1 \times {10^4}N{C^{ - 1}}{\text{ }}\left[ {\because \dfrac{e}{m}{\text{ of electron}} \simeq 1.8 \times {{10}^{11}}Ck{g^{ - 1}}} \right] \\\ \Rightarrow a = 1.8 \times {10^{15}}m/{s^2} \\\
Therefore the value of acceleration is 1.8×1015m/s21.8 \times {10^{15}}m/{s^2} .
Now we will apply the third equation of motion in order to find out the velocity as we have the acceleration, initial velocity and the distance covered.
So the third equation of motion is represented by
v2−u2=2as{v^2} - {u^2} = 2as
But here initial speed of the electron is zero i.e. u=0m/su = 0m/s
Putting the values of initial velocity, acceleration and distance, let us find the velocity of the electron.
∵v2−u2=2as v2−02=2(1.8×1015m/s2)(2×10−2m) v2=2×1.8×2×1015×10−2(m2/s2) v2=7.2×1013(m2/s2) v=7.2×1013(m2/s2) v=8.4853×106m/s v≃8.5×106m/s  \because {v^2} - {u^2} = 2as \\\ {v^2} - {0^2} = 2\left( {1.8 \times {{10}^{15}}m/{s^2}} \right)\left( {2 \times {{10}^{ - 2}}m} \right) \\\ {v^2} = 2 \times 1.8 \times 2 \times {10^{15}} \times {10^{ - 2}}\left( {{m^2}/{s^2}} \right) \\\ {v^2} = 7.2 \times {10^{13}}\left( {{m^2}/{s^2}} \right) \\\ v = \sqrt {7.2 \times {{10}^{13}}\left( {{m^2}/{s^2}} \right)} \\\ v = 8.4853 \times {10^6}m/s \\\ v \simeq 8.5 \times {10^6}m/s \\\
Hence the velocity of electron is 8.5×106m/s8.5 \times {10^6}m/s
So, the correct answer is option D.

Note- In order to solve such problems students must remember the basic relation between mass, force and acceleration. Also students must remember the laws of motion as they are very important in solving such problems and are used almost everywhere. In the presence of a uniform electric field the charged particle like an electron experiences an electrostatic force, which is responsible for the acceleration of the charged particle.