# A proton with an initial speed of 800,000 m/s is brought to rest by an electric field. a. What...

## Question:

A proton with an initial speed of 800,000 m/s is brought to rest by an electric field.

a. What was the potential difference that stopped the proton?

b. What was the initial kinetic energy of the proton, in electron volts?

## Kinetic Energy of the Proton

The change in kinetic energy of the charged particle is proportional to the potential difference.

{eq}\Delta K = 0 - \dfrac 12 mv^2 \\ \ = \ - q \Delta V {/eq}

The mass of the proton

{eq}m_p = 1.67 \times 10^{-27} kg \\ the \ charge \ of \ the \ proton\\ q = + 1.6 \times 10^{-19} \ C {/eq}

## Answer and Explanation:

Given:

{eq}m_p = 1.67 \times 10^{-27} kg \\ the \ charged \ of \ the \ proton\\ q = + 1.6 \times 10^{-19} \ C \\ v = 800,000 \ \dfrac ms {/eq}

a...

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