A positron moving from a Charged Plate to Another

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Homework Help Overview

The discussion revolves around the motion of a positron between charged plates, focusing on the concepts of kinetic and electrical energy, potential differences, and energy conservation. Participants explore the relationships between potential energy, kinetic energy, and voltage in the context of the positron's movement.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations of energy at different points (K and L) and the implications of potential differences between points (M, K, and L). Questions arise about the definitions of potential difference versus potential gain, and how to deduce voltage differences from given information.

Discussion Status

There is ongoing exploration of the relationships between potential energy and kinetic energy, with some participants providing hints and guidance on deducing potential differences from the diagram. Multiple interpretations of potential differences are being considered, and participants are questioning their understanding of the concepts involved.

Contextual Notes

Participants note that the potential difference between the charged plates is uniform, and there is some uncertainty regarding the specific values of potential at various points. The original poster expresses confusion about the differences between potential difference and potential gain, indicating a need for clarification on these concepts.

FiskiranZeka
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A positron moving from a Charged Plate to Another [SOLVED]

Can you solve the question a and b and explain it to me ?

What i know ? ( I'm not sure though );
Kinetic Work = Kinetic Energy = 1/2 m . v^2
Electrical Work = Electrical Energy = q.V

So at the begin, the positron isn't moving,
so it has Potantial energy maybe...

At the point K its Energy;
3V.q.(1/3d).d = V.q (?)

At the point L its Energy;
Vq - 4V.q.(1/2d).d = -Vq (?) ( How can energy be negative ? Oops ? )
a)EnergyK/EnergyL = ?
b) speedK / speedL = ?
 

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What's the potential difference (volts) between M and K? When the positron goes from M to K, how much energy does it gain?

Use similar thinking between M and L.
 
Doc Al said:
What's the potential difference (volts) between M and K? When the positron goes from M to K, how much energy does it gain?

Use similar thinking between M and L.
The Potential difference between the start and K isn't given in the question.
I don't know too...
But i have found the energy gain between the start and K ;
q.3V.(1/3d).d =qV


The energy gain between start and L is;
3qV gain (between start and Y Plate) + 2 qV Loss ( between Y plate and K) = 1 qV Net Gain
Start --> K = qV gain
Start --> L = qV gain

The electrical gain means, the kinetical gain,
a)
Total_Kinetic_At_K / Total_Kinetic_At_L = qV / qV = 1...
b)
Speed is related with kinetical Energy; The positron has same kinetical energy at both points, so has Same Speed at both points... Speed_K/ Speed_L = 1

Can you please check ? All right ? Any mistake ?

if all above in this message are right :) Then i have a question to you;
How you calculate the Voltage Difference between the Start and K ? ( between X plate and K point )
 
FiskiranZeka said:
The Potential difference between the start and K isn't given in the question.
Sure it is. You should be able to deduce it from the diagram. Hint: Between X and Y is 3 volts. (The field between the plates is uniform.)
 
Doc Al said:
Sure it is. You should be able to deduce it from the diagram. Hint: Between X and Y is 3 volts. (The field between the plates is uniform.)
So between X and K is 1 Volt ?
Or between X and K is still 3 Volts ?

I think;
Anywhere between X and Y plates are under 3 Volt potential difference...
But the Potential gain is depended with how much you have moved between X and Y...
If you move from X to K then you gain 1 Volt... Or if you move from Y to K then you lose 2 Volts.So; we can't say that the potential difference between X to K is 1 volts right ?
Or can we say ? Exactly I am not sure about the difference between;
Potential Difference ( Voltage Difference ) and Potential Gain ( Electrical Gain - Voltage Gain)

1) Please, clean my mind about the above statement.
2) Did i solve the problem right ?
 
FiskiranZeka said:
So between X and K is 1 Volt ?
Or between X and K is still 3 Volts ?
Between X and Y is 3 volts. Since K is 1/3 the distance between the plates, the voltage between X and K is 1 Volt. (Note that X is 3 volts above Y.)

I think;
Anywhere between X and Y plates are under 3 Volt potential difference...
But the Potential gain is depended with how much you have moved between X and Y...
If you move from X to K then you gain 1 Volt... Or if you move from Y to K then you lose 2 Volts.So; we can't say that the potential difference between X to K is 1 volts right ?
Or can we say ? Exactly I am not sure about the difference between;
Potential Difference ( Voltage Difference ) and Potential Gain ( Electrical Gain - Voltage Gain)
How about this. Let's call the voltage of plate Y to be zero (the reference point is arbitrary). So, the potential of X is +3, the potential of K is +2, the potential of Y is 0, the potential of L is +2, and the potential of Z is +4. When the positron moves from X to K, it loses PE and gains KE in the amount of 1 eV. (What's the drop in PE from X to L?)
 
When the positron moves from X to K, it loses PE and gains KE in the amount of 1 eV. (What's the drop in PE from X to L?)
Sorry, English isn't my native language.
What means eV. ?
I think its shortening of something but i don't know...

(What's the drop in PE from X to L?)
From X to L,
PE Drops 3, gains 2 --> PE Drops 1 Volt

From X to K,
PE dropps 1Volt and KE gains it somehow. ( Gains it as speed ( V^2 ) )

And yes i have exactly understood what you said except the last sentence i quoted.
Thank you so much :)
 
FiskiranZeka said:
What means eV. ?
I think its shortening of something but i don't know...
eV stands for electron-Volt, a unit of energy. See: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ev.html#c2"


From X to L,
PE Drops 3, gains 2 --> PE Drops 1 Volt
Yes.

From X to K,
PE dropps 1Volt and KE gains it somehow. ( Gains it as speed ( V^2 ) )
Yes.

(Energy is conserved. When the particle loses electric PE, it gains KE.)
 
Last edited by a moderator:
Understood everything...
Thank you very much...
I have no more questions related to this...

If you have any question for me, you are welcome.

//Problem Solved.
 

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