A horizontally-projected beam of electrons travelling at 55 km/s travels between two parallel plates separated by 7.4 cm,and 12 cm long,between which there is an electric field of 7.2 mV/m upward.If the beam enters the region at a height of 3.7 cm above the lower plate,at what height above the lower plate is the beam when it exits? The electron mass is 9.11 $\times$ 10^-31 kg. A) 0.9 cm B) 2.3 cm C) 5.1 cm D) 4.6 cm

Question

Quizplus · Accepted Answer

First, we need to find the time it takes for the electron to travel between the plates:

$d = v_{x}t$
$t = \frac{d}{v_{x}} = \frac{0.074	ext{ m}}{55	imes10^{3}	ext{ m/s}} = 1.345	imes10^{-6}	ext{ s}$

During this time, the electric field will accelerate the electrons upward. The acceleration due to the electric field is given by:

$a = \frac{F}{m} = \frac{qE}{m} = \frac{-1.602	imes10^{-19}	ext{ C}	imes7.2	imes10^{-3}	ext{ V/m}}{9.11	imes10^{-31}	ext{ kg}} \approx -1.267	imes10^{18}	ext{ m/s}^{2}$

Now we can use the kinematic equation:

$\Delta y = v_{0}t + \frac{1}{2}at^{2}$

where $\Delta y$ is the change in height of the electron (final height - initial height), $v_{0}$ is the initial velocity in the y-direction (zero), and $a$ is the acceleration due to the electric field.

$\Delta y = \frac{1}{2}at^{2} = \frac{1}{2}(-1.267	imes10^{18}	ext{ m/s}^{2})(1.345	imes10^{-6}	ext{ s})^{2} = -1.214	imes10^{-3}	ext{ m}$

Since the acceleration is upward and the initial height is 3.7 cm above the lower plate, the final height above the lower plate is:

$y_{f} = 3.7	ext{ cm} - 1.214	imes10^{-3}	ext{ m} \approx \boxed{	extbf{(B) }2.3	ext{ cm}}$

A Horizontally-Projected Beam of Electrons Travelling at 55 Km/s Travels ×\times×

A Horizontally-Projected Beam of Electrons Travelling at 55 Km/s Travels $\times$