A proton traveling at speed of 1.0 * 10⁶ m/s enters the gap between the plates of a 2.0-cm-wide parallel-plate capacitor along x-axis.The surface charge densities on the plates are $\sigma$ = $\pm$1.0 * 10^-6 C/m².How far has the proton been deflected sideways ($\Delta$y)when it reaches the far edge of the capacitor? Assume the electric field is uniform inside the capacitor and zero outside.

Question

Quizplus · Accepted Answer

First, we can calculate the electric field between the plates using the equation:
E = σ/ε0 = (1.0 × 10^-6 C/m^2) / (8.85 × 10^-12 C^2/Nm^2) = 1.13 × 10^5 N/C

Next, we can use the equation for the force experienced by a charged particle in an electric field:
F = qE
where q is the charge of the particle and E is the electric field.

The proton has a charge of 1.6 × 10^-19 C, so the force experienced by the proton is:
F = (1.6 × 10^-19 C)(1.13 × 10^5 N/C) = 1.81 × 10^-14 N

This force causes the proton to accelerate sideways, so we can use the kinematic equation:
y = (1/2)at^2
where y is the distance the proton is deflected, a is the acceleration of the proton, and t is the time it takes for the proton to cross the capacitor.

The time it takes for the proton to cross the capacitor can be found using:
d = vt
where d is the distance between the plates of the capacitor (2 cm = 0.02 m), v is the velocity of the proton, and t is the time it takes for the proton to cross the capacitor.

The velocity of the proton can be found using the relativistic formula:
v = c*sqrt(1 - (m0c^2 / (mc^2 + K))^2)
where c is the speed of light, m0 is the rest mass of the proton, m is the relativistic mass of the proton (which depends on its velocity), and K is the kinetic energy of the proton (which can be found using the classical formula K = (1/2)mv^2).

Since the proton is traveling at nearly the speed of light, we need to use the relativistic formula to find its velocity. However, since the kinetic energy of the proton is much smaller than its rest energy, we can use the approximation:
v ≈ c(1 - K / (2m0c^2))

Plugging in the numbers, we get:
K = (1/2)(1.67 × 10^-27 kg)(1.0 × 10^10 m/s)^2 = 8.35 × 10^-11 J
v ≈ (3.00 × 10^8 m/s)(1 - 8.35 × 10^-11 J / (2(1.67 × 10^-27 kg)(3.00 × 10^8 m/s)^2)) ≈ 2.99999999999983 × 10^8 m/s

Using d = vt, we can find t:
t = d / v = (0.02 m) / (2.99999999999983 × 10^8 m/s) ≈ 6.67 × 10^-11 s

Now we can use y = (1/2)at^2 with a = F/m and m = 1.67 × 10^-27 kg (the mass of the proton):
y = (1/2)(F/m)t^2 = (1/2)((1.81 × 10^-14 N) / (1.67 × 10^-27 kg))(6.67 × 10^-11 s)^2 ≈ 2.17 mm

Therefore, the proton is deflected sideways by about 2.17 mm when it reaches the far edge of the capacitor. The answer is B.

A Proton Traveling at Speed of 1 σ\sigmaσ = ±\pm± 10 * 10-6 C/m2

A Proton Traveling at Speed of 1 $\sigma$ = $\pm$ 10 * 10^-6 C/m²