What Is the Potential Difference When a Proton's Speed Increases?

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Homework Help Overview

The discussion revolves around a problem involving the potential difference experienced by a proton as it moves between two points under the influence of an electrostatic force. The proton's initial and final speeds are given, and participants are tasked with determining the potential difference based on these conditions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic energy and potential energy, referencing the equation for change in energy related to charge and potential difference. Some express confusion over the possible answers and the impact of unit conversions on their calculations.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the validity of their results. Some have found approximate answers but are uncertain about which option is correct. There is a focus on understanding the implications of the proton's speed change on potential energy and potential difference.

Contextual Notes

Participants note potential issues with unit conversions and rounding, as well as the ambiguity in the problem statement regarding the correct answer choice. There is an emphasis on the need to clarify assumptions about energy changes in conservative forces.

suldaman
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Potential difference question...

Homework Statement



A proton (mass = 1.67 10-27 kg, charge = 1.60 10-19 C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 60 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference Image .

Homework Equations


Vb-Va = 0


The Attempt at a Solution


a) +15v
b) -15V
c) -33V
d) +33V
e) -20V


im stumped with this one...any help would be appreciated.
 
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delE = q(delV)

OR

Change in E(energy) = charge*change in V(potential)

for starters
 
actually, I don't get near any of the answers there. Don't take my word for it, let me look into it more
 
ok, I did get within .4 of one of the answers there (i wasn't paying attention to units when I calculated last time)

so that way should work, methinx.
 
Pythagorean said:
ok, I did get within .4 of one of the answers there (i wasn't paying attention to units when I calculated last time)

so that way should work, methinx.

i get within .4 of the answers a well

i guess they want you to round them off

now decide which is right... a or b
 
stunner5000pt said:
i get within .4 of the answers a well

i guess they want you to round them off

now decide which is right... a or b

if you did it right, you should get either a or b, not both. They are 30 units apart.
 
Pythagorean said:
if you did it right, you should get either a or b, not both. They are 30 units apart.
I get point b about -14.6 volts relative to point a. Can't tell from the question whether that means answer a or b is correct. Probably b :rolleyes:
 
I didn't bother doing any calculations, but I can tell you the sign of the answer. Think of it this way: the final velocity was greater than the initial velocity, right? So the particle's kinetic energy increased. It moved under the influence of a conservative force solely, so the increase in kinetic energy came at the expense of a(n) ______________(increase/decrease) in potential energy. As a result, considering that the charge is positive, its final electric potential is ______________ (larger/smaller) than its initial electric potential. Therefore, the change in electric potential between the two points (final minus initial) is ______________ (positive/negative).

Fill in the blanks...:-p :devil:
 

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