Answer :
Answer:
Part(a): the acceleration of the proton is [tex]\bf{5.8 \times 10^{10}~m~s^{-2}}[/tex].
Part(b): the time taken by the proton to achieve the given velocity is [tex]\bf{24.1~\mu~s}[/tex].
Part(c): the distance travelled by the proton is [tex]\bf{16.84~m}[/tex].
Part(d): the kinetic energy of the proton is [tex]\bf{1.57 \times 10^{-15}~J }[/tex].
Explanation:
We know whenever a charge 'q' is placed in a uniform electric field, it will experience a force (F) of magnitude F = qE.
Part(a):
If 'a' be the acceleration gained by the proton when it is placed in the electric field 'E' and if '[tex]m_{p}[/tex]' be the mass of the proton, then
[tex]&& m_{p}~a = F = q~E\\&or,& a = \dfrac{q~E}{m_{p}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
Given, [tex]E = 610~N~C^{-1}[/tex] and we know that the mass of proton [tex]m_{p} = 1.67 \times 10^{-27}~Kg[/tex]. From equation(1)
[tex]a = \dfrac{1.6 \times 10^{-19}~C \times 610~N~C^{-1}}{1.67 \times 10^{-27}~Kg} = 5.8~ms^{-2} = 5.8 \times 10^{10}~m~s^{-2}[/tex]
Part(b):
If 'v' be the final velocity of the proton and 't' be the time required to this velocity, then, as the proton starts from rest,
[tex]&& v = a~t\\&or,& t = \dfrac{v}{a} = \dfrac{1.4 \times 10^{6}~m~s^{-1}}{5.8 \times 10^{10}~m~s^{-2}} = 2.41 \times 10^{-5}~s = 24.1~\mu~s[/tex]
Part(c):
If the proton travels a distance 'S' with acceleration a = [tex]5.8 \times 10^{10}~m~s^{-2}[/tex] in time t = [tex]24.1 \mu~s[/tex], then the proton will move by a distance of S, given by
[tex]S = \dfrac{1}{2}~a~t^{2} = \dfrac{1}{2} \times 5.8 \times 10^{10}~m~s^{-2} \times (2.41 \times 10^{-5})^{2}~s^{2} = 16.84~m[/tex]
Part(d):
Kinetic energy (K) of a particle of mass 'm', moving with a velocity 'v' is given by
[tex]K = \dfrac{1}{2}m v^{2}[/tex]
Here, [tex]m_{p} = 1.67 \times 10^{-27}}~kg~and~v = 1.4 \times 10^{6}~m~s^{-1}[/tex].
So, the kinetic energy of the proton at later time is
[tex]K = \dfrac{1}{2} \times 1.67 \times 10^{-27}~Kg \times (1.4 \times 10^{6})^{2}~m^{2}~s^{-2} = 1.57 \times 10^{-15}~J[/tex]
