A horizontally projected beam of protons travels at 55 km/s into the space between two horizontal, parallel plates. The plates are separated by 7.4 cm, and they are 12 cm long. There is a uniform electric field between the plates of unknown strength. The beam enters the region exactly midway (vertically) between the plates and exits the region 7.2 microns closer to the lower plate. If it were a beam of electrons instead, everything else being the same, at what height above the lower plate would the beam exit the region between the plates? The proton mass is 1.67 × 10^-27 kg, and electron mass is 9.11 × 10^-31 kg. A) 0.9 cm B) 5.0 cm C) 2.3 cm D) 4.6 cm

Question

Quizplus · Accepted Answer

We can use the equation for the deflection of charged particles in a uniform electric field:

$\Delta y = \frac{1}{2} \frac{q}{m} E \left(\frac{L}{v_{x}}ight)^2$

where:

$\Delta y$ is the vertical deflection of the beam
$q$ is the charge of the particle
$m$ is the mass of the particle
$E$ is the electric field strength
$L$ is the length of the plates
$v_{x}$ is the horizontal velocity of the beam

We can rearrange this equation to solve for the electric field strength:

$E = \frac{2 \Delta y}{q} \left(\frac{v_{x}}{L}ight)^2 m$

For the proton beam, plugging in the given values gives:

$E = \frac{2 (7.2 	imes 10^{-6} m)}{1.602 	imes 10^{-19} C} \left(\frac{55 	imes 10^3 m/s}{0.12 m}ight)^2 (1.67 	imes 10^{-27} kg)$

Solving this equation gives $E \approx 65,670 N/C$

For the electron beam, plugging in the same values except for the mass gives:

$E = \frac{2 \Delta y}{q} \left(\frac{v_{x}}{L}ight)^2 m = \frac{2 (7.2 	imes 10^{-6} m)}{-1.602 	imes 10^{-19} C} \left(\frac{55 	imes 10^3 m/s}{0.12 m}ight)^2 (9.11 	imes 10^{-31} kg)$

Solving this equation gives $E \approx -4.196 	imes 10^5 N/C$

The negative sign in this answer indicates that the electric field is now directed upwards instead of downwards. Using the same equation as before, but with the new electric field strength, we can solve for the vertical deflection of the electron beam:

$\Delta y = \frac{1}{2} \frac{q}{m} E \left(\frac{L}{v_{x}}ight)^2 = \frac{1}{2} \frac{-1.602 	imes 10^{-19} C}{9.11 	imes 10^{-31} kg} (-4.196 	imes 10^5 N/C) \left(\frac{0.12 m}{55 	imes 10^3 m/s}ight)^2$

Solving this equation gives $\Delta y \approx 5.0 	imes 10^{-2} m$, or 5.0 cm above the lower plate. Therefore, the answer is B.

A Horizontally Projected Beam of Protons Travels at 55 Km/s