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Chapter 37: Q. 18 (page 1082)

A parallel-plate capacitor with a 1.0mmplate separation is charged to 75V. With what kinetic energy, in eV, must a proton be launched from the negative plate if it is just barely able to reach the positive plate?

Short Answer

Expert verified

The kinetic energy is75eV.

Step by step solution

01

Given information

We have given that the parallel-plate capacitor with a 1.0mmplate separation is charged to75V

We need to find just barely able to reach the positive plate.

02

Explanation

Using the conservation of energy, the energy is equal to the energy if a proton starts accelerating from the positive plate in 75Vdifference, which is 75eV. So the proton should be launched with 75eVof kinetic energy so that it barely touched the positive plate.

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Most popular questions from this chapter

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