Question

A parallel-plate capacitor has 3.5 cm × 3.5 cm electrodes with surface charge densities ±1.0 × 106 C/m2. A proton traveling parallel to the electrodes at 1.9 × 106 m/s enters the center of the gap between them. By what distance has the proton been deflected sideways when it reaches the far edge of the capacitor? Assume the field is uniform inside the capacitor and zero outside the capacitor.

Answer 1

Answer:

x = 1.1*10^-7 m

Explanation:

You can consider that the electron travels in the horizontal positive x direction. Furthermore, you can consider that the electric field between the plates is totally vertical and points upward (which means that the negative plates is above the line of motion of the proton).

In order to calculate the vertical distance traveled by the electron when it has passed the region between the plates, you take into account the acceleration experienced by the proton, due to the electric field between the plates.

You first calculate the electric field between the plates, by using the following formula:

$E=\frac{\sigma}{\epsilon_o}$        (1)

σ: surface charge density of each plate = ±1.0*10^6 C/m^2

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

You replace the values of the parameters in the equation (1):

$E=\frac{1.0*10^6C/m^2}{8.85*10^{-12}C^2/Nm}=1.12*10^{17}\frac{N}{C}$

Next, you calculate the acceleration of the electron produced by the electric force. You use the second Newton law for the electric force:

$F_e=qE=ma$       (2)

q: charge of the electron = 1.6*10^-19C

m: mass of the proton = 1.67*10^-27 kg

You solve the equation (2) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19})(1.12*10^{17}N/C)}{1.67*10^{-27}kg}\\\\a=1.07*10^{25}\frac{m}{s^2}$

Next, you calculate the time that proton takes to reach the negative capacitor. You use the following formula:

$y=v_{oy}t+\frac{1}{2}at^2$     (3)

y: vertical distance from the center of the capacitor to the negative plate = 3.5cm/2 = 1.75cm = 0.0175m

voy: vertical initial speed of the proton = 0m/s (the proton enter to the region of electric field horizontally)

You solve the equation (3) for t:

$y=\frac{1}{2}at^2\\\\t=\sqrt{\frac{2y}{a}}=\sqrt{\frac{2(0.0175m)}{1.05*10^{25}m/s^2}}=5.77*10^{-14}s$

Next, you calculate the horizontal distance traveled by the proton in the time calculated. You use the following formula:

$x=v_ot$         (4)

x: length of the capacitor = 3.5cm = 0.035m

vo: initial speed of the proton = 1.9*10^6 m/s

$x=(1.9*10^6m/s)(5.77*10^{-14}s)=1.1*10^{-7}m$

The horizontal distance traveld by the electron when it impacts the negative plate of teh capacitor is 1.1*10^-7m