Question

An electron of mass 9.11 1031 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to 7.00 105 m/s in a distance of 5.00 cm. Assuming its acceleration is constant, (a) determine the force exerted on the electron and (b) compare this force with the weight of the electron, which we ignored.

Answer 1

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Initial speed of the electron, $u=3\times 10^5\ m/s$

Final speed of the electron, $v=7\times 10^5\ m/s$

Distance, d = 5 cm = 0.05 m

(a) The acceleration of the electron is calculated using the third equation of motion as :

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{(7\times 10^5)^2-(3\times 10^5)^2}{2\times 0.05}$

$a=4\times 10^{12}\ m/s^2$

Force exerted on the electron is given by :

$F=m\times a$

$F=9.11\times 10^{-31}\times 4\times 10^{12}$

$F=3.64\times 10^{-18}\ N$

(b) Let W is the weight of the electron. It can be calculated as :

$W=mg$

$W=9.11\times 10^{-31}\times 9.8$

$W=8.92\times 10^{-30}\ N$

Comparison,

$\dfrac{F}{W}=\dfrac{3.64\times 10^{-18}}{8.92\times 10^{-30}}$

$\dfrac{F}{W}=4.08\times 10^{11}$

Hence, this is the required solution.