Proton with an initial speed of 8 m/s is brought to rest by an electric field.

part a- did the proton move into a region of higher potential or lower potential?

part b - what was the potential difference that stopped the proton?
δu =

part c - what was the initial kinetic energy of the proton, in electron volts?
ki =

best answer

A) Into a region of higher potential

Explanation:

Let's remind that:

- Like charges repel each other

- Unlike charges attract each other

Here we have a proton, which is a positive charge, which is brought to rest by an electric field. This means that the electric field has slowed down the proton: so, the force exerted by the electric field on the proton was opposite to the direction of motion of the proton. But the lines of an electric field go from points at higher potential to points at lower potential - this means that the proton was actually moving towards a point at higher potential. (for example, it was moving towards another positive charge source of the field, so the potential increases as the proton approaches the source charge).

B) 3,338 V

The initial kinetic energy of the proton is given by:

where

is the proton mass

is the initial speed

Substituting,

When the proton is brought to rest, all this energy is converted into electric potential energy, given by

where

is the proton charge

is the potential difference

Since , we can solve to find the potential difference:

C) 3,338 eV

We already found the initial kinetic energy of the proton in part B), and it is given by

Now we want to convert it into electron volts; keeping in mind the conversion factor between eV and Joules,

we find:

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

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