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# Potential Energy of a +ve Charge. watch

1. I know that the potential energy will be equal to Fs, but I think that it will increase by Fs because it's moving away from positive plate automatically. We don't have to do work on it. So its potential energy should increase. Doesn't potential energy decrease when we've to do some work to bring it to the plate which repels that particular charge?
Mark scheme's answer is A. I think B should be the answer according to my explanation above!
2. Electric field lines always go from positive to negative, and in the case of a positive test charge if you push it in the direction of the E-field you're not doing any work against the field.

The field lines go from higher potential to lower potential. That means the potential should decrease.
3. (Original post by trm90)
Electric field lines always go from positive to negative, and in the case of a positive test charge if you push it in the direction of the E-field you're not doing any work against the field.

The field lines go from higher potential to lower potential. That means the potential should decrease.
Oh, so as Potential = Work done/Charge that's why as potential will decrease potential energy will also decrease, right?
(Original post by Stonebridge)
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5. (Original post by Zishi)
Oh, so as Potential = Work done/Charge that's why as potential will decrease potential energy will also decrease, right?
It's FS. It doesn't matter now you move through the field, it just matters the distance travelled along it, if that makes sense. So the same work is done moving to y as it would be moving straight to the right. You could even do loop-the-loops before you got there and the change in potential would be the same.

_Kar.
6. (Original post by Kareir)
It's FS. It doesn't matter now you move through the field, it just matters the distance travelled along it, if that makes sense. So the same work is done moving to y as it would be moving straight to the right. You could even do loop-the-loops before you got there and the change in potential would be the same.

_Kar.
I know it's FS, but I don't understand why the potential energy will decrease?
7. (Original post by Zishi)
I know it's FS, but I don't understand why the potential energy will decrease?

As you have correctly said, Electric Potential is equal to the work done per unit charge....

You also know that field lines go AWAY from a positive charge? And thus the electric potential due a positive charge is positive...

You know that work needs to be DONE on a charge to move it from X to Y...

dW(on charge) = q dV

therefore, the charge DOES negative work?

dW(by charge) = - q dV

W = F s

F s = - q dV

dV = - (Fs / q )

Therefore... it is a decrease
8. on thinking again... its better to think of the work done in moving a charge between two points as the POTENTIAL DIFFERENCE... this is what everyone is most familiar with...

im sure you should be able to state the difference by EMF and PD in terms of a electrical circuit... namely that the PD between two points in a circuit is the amount of electrical potential energy converted to other forms of energy when unit charge passes from one point to another... EMF is simply the transfer of other forms of energy to electrical potential energy.... u need to think in terms of PD.....
9. (Original post by chemdweeb1234)
As you have correctly said, Electric Potential is equal to the work done per unit charge....

You also know that field lines go AWAY from a positive charge? And thus the electric potential due a positive charge is positive...

You know that work needs to be DONE on a charge to move it from X to Y...

dW(on charge) = q dV

therefore, the charge DOES negative work?

dW(by charge) = - q dV

W = F s

F s = - q dV

dV = - (Fs / q )

Therefore... it is a decrease

(Original post by chemdweeb1234)
on thinking again... its better to think of the work done in moving a charge between two points as the POTENTIAL DIFFERENCE... this is what everyone is most familiar with...

im sure you should be able to state the difference by EMF and PD in terms of a electrical circuit... namely that the PD between two points in a circuit is the amount of electrical potential energy converted to other forms of energy when unit charge passes from one point to another... EMF is simply the transfer of other forms of energy to electrical potential energy.... u need to think in terms of PD.....
Hmm, I have one more way of explaining it - is it correct?: As the positive charge is repelled by the positive plate to move towards Y, so its speed increases and according to law of conservation of energy as speed increases, the Electric potential energy must decrease!
10. (Original post by Zishi)
Hmm, I have one more way of explaining it - is it correct?: As the positive charge is repelled by the positive plate to move towards Y, so its speed increases and according to law of conservation of energy as speed increases, the Electric potential energy must decrease!
Bingo :-) its what i said above... potential difference... EPE converted to KE
11. (Original post by Zishi)
I think it already has been, but if ever you can't decide whether something gains or loses PE just do the following.
Turn the diagram on its side and imagine it is a gravitational field with a mass m instead of a charge.

The gravitational force acts downwards and the arrows represent the field.
If the mass moves from X to Y does it gain or lose potential energy mgh?
The answer is obvious, I think.
Hope this helps.
12. (Original post by Stonebridge)
I think it already has been, but if ever you can't decide whether something gains or loses PE just do the following.
Turn the diagram on its side and imagine it is a gravitational field with a mass m instead of a charge.

The gravitational force acts downwards and the arrows represent the field.
If the mass moves from X to Y does it gain or lose potential energy mgh?
The answer is obvious, I think.
Hope this helps.
thats the most brilliant way of thinking about it... you legend.
13. (Original post by Stonebridge)
I think it already has been, but if ever you can't decide whether something gains or loses PE just do the following.
Turn the diagram on its side and imagine it is a gravitational field with a mass m instead of a charge.

The gravitational force acts downwards and the arrows represent the field.
If the mass moves from X to Y does it gain or lose potential energy mgh?
The answer is obvious, I think.
Hope this helps.
Wow, thanks. Then just two more things: Can we always take gravitational field in the direction of electric field in questions like this? If the question asked about 'electrical' potential energy - would this have given a correct answer?
14. would this have given a correct answer?
No. What I wrote is not an answer to the question. It was a way of remembering how to work it out.
The point is that * if an object moves as a result of the force applied to it by a field, it loses PE.
This is true both for gravitational and electrical PE. Most people find it easier to think in terms of gravity as its more familiar. The gravitational field is always attractive though, while electric fields can both attract and repel. So you need to be careful. There is no subsitute for learning the theory.

* And, if you move an object against the force applied to it by the field, the object gains PE. (Same as in a gravitational field when you lift something up.)

That is the rule you have to learn.
15. (Original post by Stonebridge)
No. What I wrote is not an answer to the question. It was a way of remembering how to work it out.
The point is that * if an object moves as a result of the force applied to it by a field, it loses PE.
This is true both for gravitational and electrical PE. Most people find it easier to think in terms of gravity as its more familiar. The gravitational field is always attractive though, while electric fields can both attract and repel. So you need to be careful. There is no subsitute for learning the theory.

* And, if you move an object against the force applied to it by the field, the object gains PE. (Same as in a gravitational field when you lift something up.)

That is the rule you have to learn.
I've learned that rule now - thank you!

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