Answer to Question #138235 in Mechanics | Relativity for Harsd

Question

An electron moving parallel to the x-axis has an initial speed of 3.70 x 106 m/s at the origin. Its speed is reduced to 1.40 x 105 m/s at the point x=2.00 cm. (a) Calculate the electric potential difference between the origin and that point. (b) Which point is at the higher potential?

Answer 1

Solution

a) the potential difference is given by

$\Delta V=-\frac{m_e (v_f^2-v_i^2) }{2q}$

$\Delta V\\=-\frac{9.1\times10^{-31}((1.4\times10^5)^2- (3.7\times10^6)^2 ) }{2(-1.6\times10^{-19}) }$

$\Delta V=-38.9V$

b) $\Delta V$ is negative it's means potential is higher at origin.