Answer to Question #142586 in Electric Circuits for qwetdfg

Question

An electron moving parallel to the x axis has an initial speed of 3.7 x 10^6 m/s at the origin. Its
speed is reduced to 1.4 x 10^5 m/s at x = 2 cm. Calculate the electric potential difference
between the origin and x -= 2 cm.

Answer 1

Since the change in kinetic energy is equal to the product of the potential difference per particle charge, then

$\Delta E_c=q\times U$ ;

$U=\frac{\Delta E_c}{q}=\frac{\frac{m}{2}\times(v_1^2-v_2^2)}{q}$ $=\frac{\frac{9.31\times10^{-31}}{2}\times((3.7\times10^6)^2-(1.4\times10^5)^2)}{1.6\times10^{-19}}=39.77$ V.