# Homework Help: Electric potential and velocity of a proton

1. Oct 27, 2008

### Foxhound101

1. The problem statement, all variables and given/known data
http://session.masteringphysics.com/problemAsset/1020825/7/jfk.Figure.21.P56.jpg
A proton's speed as it passes point A is 4.40×10^4 m/s. It follows the trajectory shown in the figure.

What is the proton's speed at point B?

2. Relevant equations

Knowing this would be very helpful

3. The attempt at a solution

No attempt yet...If I knew what formulas to use, that would be very helpful.

2. Oct 27, 2008

### cepheid

Staff Emeritus
By definition, an electric field is accelerating the charge in the direction of decreasing electric potential. Hence, the charge's kinetic energy (and therefore velocity) is increasing.

But you don't know the electric field and therefore the force. So how do you figure out the change in kinetic energy? Because you know the potential, which is equivalent information. Think about it in terms of conservation of energy. The proton loses electric potential energy in moving from point A to B, which means that it must gain exactly that amount of kinetic energy. And you know, from the diagram, exactly how much potential energy per unit charge is lost in going from A to B. NOW do you understand what calculations need to be carried out?

Physics is not about memorizing formulae, it's about understanding and applying concepts. We can teach you equations, but if you don't understand physics, you won't know which ones to use in which situations. ;-)

3. Oct 27, 2008

### Foxhound101

Sadly, I still do not understand. This entire chapter confuses me.

4. Oct 27, 2008

### cepheid

Staff Emeritus
Okay. To summarize what I said in my second paragraph:

potential energy lost = kinetic energy gained.

potential energy lost is depicted on the diagram. The lines shown are lines of constant electric potential (like contours). The proton is crossing them, going "downhill" as it were. How much potential energy does it lose?

5. Oct 27, 2008

### Foxhound101

40v?

6. Oct 27, 2008

### cepheid

Staff Emeritus
Right, exactly. The potential difference is 30 V - (-10 V) = 40 V.

Now, electric potential is potential energy PER UNIT CHARGE (1 volt = 1 joule/coulomb).
So, given *that* potential difference (40 V) between points A and B, how much does the potential energy of this *particular* charge decrease?

7. Oct 27, 2008

### Foxhound101

1 v = 1 j/c

so...

40 v = 40 j/c

40 j/c

8. Oct 27, 2008

### cepheid

Staff Emeritus
Dude, that's how much energy is lost PER UNIT charge. I asked you for the total amount of potential energy lost (in JOULES! Energy is measured in joules). Hint, we have a certain *amount* of charge passing by, and we know how much energy is lost per UNIT of charge.

Edit: You need to review the concept of electric potential. Make sure you understand why I am saying what I am saying. Electric potential measures how much potential energy there is at a point PER UNIT charge. I.e. a "test" charge of 1 coulomb sitting at that point would have that potential energy.

9. Oct 27, 2008

### Foxhound101

Alright, thanks for the help. I shall try again later after reviewing the material.