Proton Acceleration - Do you Agree?

[TrippingSunwise]

A proton, initially at rest, is accelerated from plate A to plate B, and acquires 1.92 * 10-17 J of kinetic energy.
i. Which plate is positive and which is negative?
ii. What is the potential difference?
iii. Sketch the correct direction of E between the plates. Does the proton move with or against the field E? Does this agree with the gain (or loss) calculated in ii?? Explain?

My Solution:

i. Plate A is positive Plate B is negative
ii. q * delta V = 1.92 * 10-17 J

Delta V = -1.92 * 10-17 / 1.602 * 10-19 C = -1.1985 * 10^2 V
The potential difference is -1.1985 * 10^2 V
This is a loss in potential

iii. The proton moves with the field and this agrees with my previous calculation as an electron moving with a field is like a mass falling downward within Earth's gravitational fieldd. There is no need for an external force so there we witness a loss of potential.

Does this make sense to everyone?

 

[whozum]

That last line is a bit confusing.

In (ii) the potential between the plates isn't affected by the proton's gain in kinetic energy. The proton's final kinetic energy is due to the presence of the potential difference. The proton experienced a change in PE when flowing from A to B, this change in PE results in the change in KE.

 

[Dr.Brain]

1)Ok for your first question, as the proton gains energy moving from A to B ... B should have a higher potential, that is B should be positive.

2) Potential difference can be calculated:

$qV= 1.92 x 10^-17$

Calculate V

3)Electric field lines go from +ve to -ve plates and the proton moves against the field.

The proton has to do work in moving against the field ...so that much energy is stored in it which is the gain in energy.

 

[OlderDan]

1)Ok for your first question, as the proton gains energy moving from A to B ... B should have a higher potential, that is B should be positive.

2) Potential difference can be calculated:

$qV= 1.92 x 10^-17$

Calculate V

3)Electric field lines go from +ve to -ve plates and the proton moves against the field.

The proton has to do work in moving against the field ...so that much energy is stored in it which is the gain in energy.

The OP has it right. Plate A is positive and plate B is negative. If they were reversed, the proton would not accelerate toward plate B. Plate A is at the higher potential. The proton loses potential energy as it moves from plate A to plate B, gaining kinetic energy in the process. There is a bit of a problem with the last statement.

A proton, initially at rest, is accelerated from plate A to plate B, and acquires 1.92 * 10-17 J of kinetic energy.

iii. The proton moves with the field and this agrees with my previous calculation as an electron moving with a field is like a mass falling downward within Earth's gravitational fieldd. There is no need for an external force so there we witness a loss of potential.

For gravity, the force is always in the direction of the gravitational field and in the direction of lower potential energy. For charges, the electric force direction is always toward lower potential energy, but not always toward lower electric potential. I assume the word electron was supposed to be proton, though it can be correct either way if you by "field" you mean the potentail energy field rather than the electric potential field. For a positive charge like the proton, the force is in the direction of the electric potential.