Question

A proton initially traveling at 50,000 m/s is shot through a small hole in the negative plate of a parallal-plate capacitor. The electric field strength inside the capacitor is 1,500 V/m. How far does the proton travel above the negative plate before temporarily coming to rest and reversing course? Assume the proton reverses course before striking the positive plate.

Answer 1

Answer:

x = 8.699 10⁻³ m

Explanation:

The proton feels an electric charge that is the opposite direction of speed, let's look for acceleration using Newton's second law

      F = m a

        F = q E

      a = q E / m

     

      a = 1.6 10⁻¹⁹ 1500 / 1.67 10⁻²⁷

      a = 1,437 10¹¹ m / s²

Now we can use kinematic relationships

      v² = v₀² - 2 a x

When at rest the speed is zero (v = 0)

      x = v₀² / 2 a

Let's calculate

     x = 50,000² / (2 1,437 10¹¹)

     x = 8.699 10⁻³ m