# A positron moving from a Charged Plate to Another

1. Dec 20, 2008

### FiskiranZeka

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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2. Dec 20, 2008

### Staff: Mentor

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.

3. Dec 20, 2008

### FiskiranZeka

The Potential difference between the start and K isn't given in the question.
I dont 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 )

4. Dec 20, 2008

### Staff: Mentor

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.)

5. Dec 20, 2008

### FiskiranZeka

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 im not sure about the difference between;
Potential Difference ( Voltage Difference ) and Potential Gain ( Electrical Gain - Voltage Gain)

2) Did i solve the problem right ?

6. Dec 20, 2008

### Staff: Mentor

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.)

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?)

7. Dec 20, 2008

### FiskiranZeka

Sorry, English isn't my native language.
What means eV. ?
I think its shortening of something but i dont know...

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 :)

8. Dec 20, 2008

### Staff: Mentor

eV stands for electron-Volt, a unit of energy. See: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ev.html#c2"

Yes.

Yes.

(Energy is conserved. When the particle loses electric PE, it gains KE.)

Last edited by a moderator: Apr 24, 2017
9. Dec 20, 2008

### FiskiranZeka

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.