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.

Question

Physics

Mister

27 December, 19:34

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.

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Answers (1)

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Marquis · Answer 1

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