Question

An electron moving along the x axis has an initial speed of 1 × 106 m/s at the origin. Its speed is reduced to 5 × 105 m/s at the point xP , 3 cm away from the origin. Calculate the magnitude of the potential difference between this point and the origin. The mass of the electron is 9.109 × 10−31 kg and the elemental charge is 1.602 × 10−19 C. Answer in units of V.

Answer 1

Answer:

-2.13 V

Explanation:

Given parameters are:

$v_i = 1 * 10^6$ m/s

$v_f = 5 * 10^5$ m/s

$q = -e = 1.602 * 10^{-19}$ C

$m = 9.109 * 10^{-31}$ kg

By conservation of energy principle, we know that the potential difference between two points is equal to the change in the kinetic energy between these points.

$-\Delta U = \Delta KE\\\\-e\Delta V = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2$

Hence,

$1.602*10^{-19}\Delta V = \frac{1}{2}*9.11*10^{-31}((5*10^5)^2 - (1*10^6)^2)$

⇒ $\Delta V = -2.13$ V