Answer:
[tex]3.64\times 10^{-14} N[/tex] is the magnitude of the force required to accelerate an electron of given mass.
Explanation:
Initial velocity of the electron = u
Final velocity of the electron = v = [tex]2.0\times 10^7 m/s[/tex]
Acceleration of the electron = a =?
Distance covered by electron= s = 0.50 cm = 0.005 m ( 1 cm = 0.01 m)
Using third equation of motion :
[tex]v^2-u^2=2as[/tex]
[tex](2.0\times 10^7 m/s)-0^2=2\times a\times 0.005 m[/tex]
[tex]a=4\times 10^{16} m/s^2[/tex]
Mass of an electron = [tex]m=9.1 \times 10^{-31} kg[/tex]
Force on moving electron = F
[tex]F = m\times a[/tex]
[tex]=9.1 \times 10^{-31} kg\times 4\times 10^{16} m/s^2=3.64\times 10^{-14} N[/tex]
[tex]3.64\times 10^{-14} N[/tex] is the magnitude of the force required to accelerate an electron of given mass.
The magnitude of the force required to accelerate an electron is [tex]3.64 \times 10^{-18}N[/tex]
According to Newton's second law, the formula for calculating the required force is expressed as:
Get the required acceleration;
[tex]v^2=u^2+2as\\(2.0 \times 10^7)^2=0^2+2(0.005)a\\4.0 \times 10^{14}=0.01a\\a=4.0\times 10^{12}m/s^2[/tex]
Geet the magnitude of the force required:
[tex]F=9.1\times 10^{-31} \times 4.0 \times 10^{12}\\F=36.4 \times 10^{-19}N\\F=3.64 \times 10^{-18}N[/tex]
Hence the magnitude of the force required to accelerate an electron is [tex]3.64 \times 10^{-18}N[/tex]
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