An electron with velocity v = 1.0 x 10⁶ m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m.If the distance the electron travels through the field is 1.0 cm,how far is it deviated (Y)in its path when it emerges from the electric field? (m_e = 9.1 x 10^-31 kg,e = 1.6 x 10^-19 C) A)2.2 mm B)4.4 mm C)2.2 cm D)4.4 cm

Question

Quizplus · Accepted Answer

We first need to find the acceleration of the electron due to the electric field:

a = eE/m = (1.6 x 10^-19 C)(500 V/m)/(9.1 x 10^-31 kg) = 8.79 x 10^14 m/s^2

We know the distance the electron travels through the field, so we can use the kinematic equation:

Y = (1/2)at^2 = (1/2)(8.79 x 10^14 m/s^2)(1 x 10^-2 m)^2 = 4.4 x 10^-4 m = 4.4 mm

Therefore, the electron is deviated by 4.4 mm in its path when it emerges from the electric field. The answer is (B).

An Electron with Velocity V = 1