An electron with velocity v = 2.0 × 10⁶ m/s is sent between the plates of a capacitor where the electric field is E = 200 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 × 10−³¹ kg,e = 1.6 × 10−¹⁹ C) A)0.84 mm B)0.44 mm C)0.44 cm D)0.84 cm E)1.4 mm

Question

Quizplus · Accepted Answer

The force (F) on the electron can be calculated using the equation F = qE, where q is the charge of the electron and E is the electric field strength.

F = (1.6 × 10^-19 C)(200 V/m) = 3.2 × 10^-17 N

Using the equation F = ma, where m is the mass of the electron and a is its acceleration, we can find its acceleration in the electric field.

a = F/m = (3.2 × 10^-17 N)/(9.1 × 10^-31 kg) = 3.5 × 10^13 m/s^2

The distance (Y) the electron is deviated can be calculated using the equation Y = (1/2)at^2, where t is the time it takes the electron to travel through the electric field.

t = d/v = (1.0 × 10^-2 m)/(2.0 × 10^6 m/s) = 5.0 × 10^-9 s

Y = (1/2)(3.5 × 10^13 m/s^2)(5.0 × 10^-9 s)^2 = 0.44 mm

Therefore, the best choice is B, which represents the answer of 0.44 mm.

An Electron with Velocity V = 2