Problem 11-55 Potential and kinematics in thundercloud - Part 8

Under a particular thundercloud, there is a uniform electric field of \(9.3 \times10^3\; V/m\) upward.

An electron in the field, initially at rest, travels freely for \(3.5 \times 10^{-6} \; m\) before undergoing a collision with a gas molecule. Just before it hits the molecule, what is its speed? Use energy methods.

[Ans. \(1.1 \times 10^5\; m/s\)]

Diagram A showing the field direction and the direction of motion of the electron

\(E_{K1} + U_1 = E_{K2} + U_2 \\ U = qEy \)

Correct!

\(0 + 0 = ½\; mv_2{^2} - qEy\)

Solving for \(v_2{^2}\)

 \(v_2^2 = \frac{2qEy}{m} = \frac{2(1.6 \times 10^{-19} C)(9.3 \times 10^3\; V/m)(3.5 \times 10^{-6} \;m)}{9.1 \times 10^{-31}\;kg} = 1.145 \times 10^{10} \\ v_2 = 1.1 \times 10^5 \; m/s\)
 

You have completed this problem.