Explanation:
It is given that,
Mass of an electron, [tex]m=9.11\times 10^{-31}\ kg[/tex]
Initial speed of the electron, [tex]u=3\times 10^5\ m/s[/tex]
Final speed of the electron, [tex]v=7\times 10^5\ m/s[/tex]
Distance, d = 5 cm = 0.05 m
(a) The acceleration of the electron is calculated using the third equation of motion as :
[tex]a=\dfrac{v^2-u^2}{2d}[/tex]
[tex]a=\dfrac{(7\times 10^5)^2-(3\times 10^5)^2}{2\times 0.05}[/tex]
[tex]a=4\times 10^{12}\ m/s^2[/tex]
Force exerted on the electron is given by :
[tex]F=m\times a[/tex]
[tex]F=9.11\times 10^{-31}\times 4\times 10^{12}[/tex]
[tex]F=3.64\times 10^{-18}\ N[/tex]
(b) Let W is the weight of the electron. It can be calculated as :
[tex]W=mg[/tex]
[tex]W=9.11\times 10^{-31}\times 9.8[/tex]
[tex]W=8.92\times 10^{-30}\ N[/tex]
Comparison,
[tex]\dfrac{F}{W}=\dfrac{3.64\times 10^{-18}}{8.92\times 10^{-30}}[/tex]
[tex]\dfrac{F}{W}=4.08\times 10^{11}[/tex]
Hence, this is the required solution.
