Question

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

a. Did the proton move into a region of higher potential or lower potential?

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

Short answer

a. The proton shifted to a higher potential.

b.$∆V=\mathbf{3340}\mathbf{}\mathit{V}$

Step-by-step solution

Given information and theory used  

Given : Proton's initial speed : $800,000m/s$

Theory used :

From the equation of conservation of energy, we have :$q∆V=\frac{m{v}^2}{2}$

Determining if the proton moves into a region of higher potential or lower potential 

a) In an electric field, positive charges experience an electrostatic force that points in the direction of decreasing electric potential.

This indicates that as the proton decelerated, the force pressing on it shifted in the opposite direction as the proton moved.

The proton, in other words, shifted to a higher potential.

Determining the potential difference that stopped the proton 

b. We have from the equation of conservation of energy :$q∆V=\frac{m{v}^2}{2}\Rightarrow ∆V=\frac{m{v}^2}{2q}$

In terms of numbers,

$∆V=\frac{1.67\times {10}^{-27}\times (8\times {10}^6)}{2\times 1.6\times {10}^{-19}}=\boxed{\mathbf{3340}\mathbf{}\mathit{V}}$