question_answer

In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of \[\text{5}\times \text{l}{{0}^{\text{6}}}\]m/s in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is \[\text{9}.\text{1}\times \text{1}{{0}^{-\text{31}}}\] kg.

Answer:

\[u=0,\,\,v=5\times {{10}^{6}}m/s,\,\,\,s=1cm=\frac{1}{100}m,\,\,F=?,\,\,m=9.1\times {{10}^{-31}}kg\] We know,\[F=ma\] How to get a? Applying, \[a=\frac{{{V}^{2}}-{{u}^{2}}}{2s}=\frac{(5\times {{10}^{6}})-{{0}^{2}}}{2\times {}^{1}/{}_{100}}\] \[=\frac{25\times {{10}^{12}}\times {{10}^{2}}}{2}=12.5\times {{10}^{14}}m/{{s}^{2}}\] \[\therefore F=9.1\times {{10}^{-31}}=12.5\times {{10}^{14}}N\] \[=113.75\times {{10}^{-17}}=1.1\times {{10}^{-15}}N.\]