Electric potential

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16Characters....
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#1
Report Thread starter 6 years ago
#1
Hello,

I am currently doing a physics question and am getting the exact negatives of what I should be getting. The question is "An electron in a beam is accelerated from a potential of -50V to a potential of 450V. Calculate the electric potential energy of the electron a) at a potential of -50V b) at a potential of 450V."

I did:

E_p = QV = (-1.6 \times 10^{-19})(-50) = 8.0 \times 10^{-18} J
and
E_p = QV = (-1.6 \times 10^{-19})(450) = -7.2 \times 10^{-17} J

But the answers should be  - 8.0 \times 10^{-18} J and  7.2 \times 10^{-17} J. I don't understand where I have gone wrong; which of the two quantities was I supposed to negate (and why?)?
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#2
Report 6 years ago
#2
(Original post by 16Characters....)
Hello,

I am currently doing a physics question and am getting the exact negatives of what I should be getting. The question is "An electron in a beam is accelerated from a potential of -50V to a potential of 450V. Calculate the electric potential energy of the electron a) at a potential of -50V b) at a potential of 450V."

I did:

E_p = QV = (-1.6 \times 10^{-19})(-50) = 8.0 \times 10^{-18} J
and
E_p = QV = (-1.6 \times 10^{-19})(450) = -7.2 \times 10^{-17} J

But the answers should be  - 8.0 \times 10^{-18} J and  7.2 \times 10^{-17} J. I don't understand where I have gone wrong; which of the two quantities was I supposed to negate (and why?)?

Have you posted the full question? It's quite tricky to comment on the sign of the values.
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16Characters....
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#3
Report Thread starter 6 years ago
#3
(Original post by Mehrdad jafari)
Have you posted the full question? It's quite tricky to comment on the sign of the values.
"An electron in a beam is accelerated from a potential of -50V to a potential of 450V. Calculate:
a) i) The potential energy of the electron at -50v
ii) The potential energy of the electron at 450V
b) The change in potential energy of the electron."
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#4
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#4
(Original post by 16Characters....)
"An electron in a beam is accelerated from a potential of -50V to a potential of 450V. Calculate:
a) i) The potential energy of the electron at -50v
ii) The potential energy of the electron at 450V
b) The change in potential energy of the electron."
This is merely my own understanding of it and I will be happy if anyone could provide a better explanation of the significance of the signs

When the electron is in the negative potential (-50V), it will be repelled by the electric field lines towards the zero potential, with the release of energy in the process because no external work done is required. Therefore, at thhe potential of -50V, the electron has the energy stored calculated in part a and the negative sign is an indication of the energy released (you can think of it in the same way as gravitational potential energy).
But when the electron is in the positive potential (450V), energy is required to move the electron from 450V to zero potential (zero potential is our point of reference and is considered to be at infinity), because the electron is attracted towards the positive potential. Hence, the positive sign is denoted to indicate the energy required.
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#5
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#5
(Original post by Mehrdad jafari)
This is merely my own understanding of it and I will be happy if anyone could provide a better explanation of the significance of the signs

When the electron is in the negative potential (-50V), it will be repelled by the electric field lines towards the zero potential, with the release of energy in the process because no external work done is required. Therefore, at thhe potential of -50V, the electron has the energy stored calculated in part a and the negative sign is an indication of the energy released (you can think of it in the same way as gravitational potential energy).
But when the electron is in the positive potential (450V), energy is required to move the electron from 450V to zero potential (zero potential is our point of reference and is considered to be at infinity), because the electron is attracted towards the positive potential. Hence, the positive sign is denoted to indicate the energy required.
Thank you for the answer, but I am not sure if I am following.Can you please explain what was wrong with my original logic? :
- Electric potential is the electrical potential energy per positive unit charge at a given point in a field
- Therefore a negative potential means that the force acting on a positive test charge placed at that point is attractive
- Hence an electron would be repelled
- Repulsion corresponds to positive potential energy.
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#6
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#6
(Original post by 16Characters....)
Thank you for the answer, but I am not sure if I am following.Can you please explain what was wrong with my original logic? :
- Electric potential is the electrical potential energy per positive unit charge at a given point in a field
- Therefore a negative potential means that the force acting on a positive test charge placed at that point is attractive
- Hence an electron would be repelled
- Repulsion corresponds to positive potential energy.
There was nothing wrong your your logic, you just substituted the numbers into the equation and worked out the final sign mathematically.

That's the definition of electric field strength, and not electric potential. Electric potential is the work done per unit test charge (or positive test charge) to be moved from infinity to a given point in the field. I don't know where you got that question from but at A level the the questions are given with a positive charge and a negative potential and everything would mathematically work out. But in this case, the charge is negative in two different potentials and so you would have to think logically to denote a correct sign to the potential energy.

I hope this makes quite sense now if you read my previous post again.

Edited
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#7
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#7
(Original post by Mehrdad jafari)
There was nothing wrong your your logic, you just substituted the numbers into the equation and worked out the final sign mathematically.

That's the definition of electric field strength, and not electric potential. Electric potential is the work done per unit test charge (or positive test charge) to be moved from infinity to that point. I don't know where you got that question from but at A level the the questions are given with a positive charge and a negative potential and everything would mathematically work out. But in this case, the charge is negative in two different potentials and so you would have to think logically to use denote a correct sign to the potential energy.

I hope this makes quite sense now if you read my previous post again.
The question was from the standard AQA textbook. That textbook also gave two definitions for electric potential; the work done in moving a unitary test charge from infinity to that point, and also as electrical potential energy per unit positive charge. Having asked around, this question has tripped everybody up in my class. But yes I think I do understand it now. Thanks for your time.
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#8
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#8
(Original post by 16Characters....)
Hello,

I am currently doing a physics question and am getting the exact negatives of what I should be getting. The question is "An electron in a beam is accelerated from a potential of -50V to a potential of 450V. Calculate the electric potential energy of the electron a) at a potential of -50V b) at a potential of 450V."

I did:

E_p = QV = (-1.6 \times 10^{-19})(-50) = 8.0 \times 10^{-18} J
and
E_p = QV = (-1.6 \times 10^{-19})(450) = -7.2 \times 10^{-17} J

But the answers should be  - 8.0 \times 10^{-18} J and  7.2 \times 10^{-17} J. I don't understand where I have gone wrong; which of the two quantities was I supposed to negate (and why?)?
I agree with you. An electron moving from a neutral into a negative area will have a repulsive force upon it, so will need energy to move into the negative area. This means its potential energy will increase. Thus its potential energy at -50v should be positive. Reverse argument for the positive potential.

You can just write [tex][/tex].
(Original post by Mehrdad jafari)
There was nothing wrong your your logic, you just substituted the numbers into the equation and worked out the final sign mathematically.

That's the definition of electric field strength, and not electric potential. Electric potential is the work done per unit test charge (or positive test charge) to be moved from infinity to a given point in the field.
That is not the definition of electric field strength, that's one definition of electrical potential. Your definition is equal(but better defined).
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#9
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#9
(Original post by 16Characters....)
The question was from the standard AQA textbook. That textbook also gave two definitions for electric potential; the work done in moving a unitary test charge from infinity to that point, and also as electrical potential energy per unit positive charge. Having asked around, this question has tripped everybody up in my class. But yes I think I do understand it now. Thanks for your time.

(Original post by morgan8002)
I agree with you. An electron moving from a neutral into a negative area will have a repulsive force upon it, so will need energy to move into the negative area. This means its potential energy will increase. Thus its potential energy at -50v should be positive. Reverse argument for the positive potential.

You can just write [tex][/tex].

That is not the definition of electric field strength, that's one definition of electrical potential. Your definition is equal(but better defined).
That's cool. But I would say that the electric potential energy per unit charge doesn't have much significance. The electric potential energy "stored" per unit charge would be better
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#10
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#10
(Original post by Mehrdad jafari)
That's cool. But I would say that the electric potential energy per unit charge doesn't have much significance. The electric potential energy "stored" per unit charge would be better
I don't think it makes any difference. Potential energy has to be stored so mentioning it implies it is stored somewhere.
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#11
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#11
(Original post by morgan8002)
I agree with you. An electron moving from a neutral into a negative area will have a repulsive force upon it, so will need energy to move into the negative area. This means its potential energy will increase. Thus its potential energy at -50v should be positive. Reverse argument for the positive potential.

You can just write [tex][/tex].

That is not the definition of electric field strength, that's one definition of electrical potential. Your definition is equal(but better defined).
So you are saying that you agree with my answer and disagree with the textbook?

And cheers for the LaTeX advice.
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#12
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#12
(Original post by 16Characters....)
So you are saying that you agree with my answer and disagree with the textbook?

And cheers for the LaTeX advice.
Yes. It saves a lot of typing over time.
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#13
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#13
(Original post by morgan8002)
Yes. It saves a lot of typing over time.
Well I wasted my sunday evening then :P Thanks for the help.
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#14
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#14
(Original post by morgan8002)
I don't think it makes any difference. Potential energy has to be stored so mentioning it implies it is stored somewhere.
Potential energy per unit charge doesn't suggest that a certain amount of energy is stored anywhere.
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#15
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#15
(Original post by Mehrdad jafari)
Potential energy per unit charge doesn't suggest that a certain amount of energy is stored anywhere.
It does because potential energy must be stored. If it exists then it is stored. eg. gravitational potential energy is stored by the arrangement of mass, electrical potential energy is stored by the arrangement of charge and so on.
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#16
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#16
(Original post by 16Characters....)
Well I wasted my sunday evening then :P Thanks for the help.
That's just a sunday evening. People "waste" their lives understanding something
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#17
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#17
(Original post by morgan8002)
It does because potential energy must be stored. If it exists then it is stored. eg. gravitational potential energy is stored by the arrangement of mass, electrical potential energy is stored by the arrangement of charge and so on.
That's true but gravitational potential energy isn't defined the same way though.
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#18
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#18
(Original post by Mehrdad jafari)
That's true but gravitational potential energy isn't defined the same way though.
It is.
E_p = -\int_x^{\infty} Fdx

Edit: divide by mass or charge to get potentials
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#19
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#19
(Original post by morgan8002)
It is.
E_p = -\int_x^{\infty} Fdx

Edit: divide by mass or charge to get potentials
That's not as vague as saying gravitational potential is energy per unit mass
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#20
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#20
You don't have to put the negative in front of the electron charge as you only need to use the magnitude
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