taking a negative charge from negative potential to positive potential

[thebillsfan]

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alright, so does "positive potential"/"high potential" mean an area of high positive charge density? if that's the case, why would taking a negative charge from an area of low potential to high potential require work? shouldnt the negative charge willingy go where there is more positive charge?

 

[549]

alright, so does "positive potential"/"high potential" mean an area of high positive charge density? if that's the case, why would taking a negative charge from an area of low potential to high potential require work?

W = -∆PE = -q∆V

If you are going from an area of low potential to high potential, ∆V is +. q is -, as you stated. Therefore, W will be +.

shouldnt the negative charge willingy go where there is more positive charge?

It moves toward the area of high potential due to the electric force it feels, which is accelerating it. Therefore KE will be increasing, and that's where the work is done. Think about it this way maybe: if you drop a book from a height, it wants to go to the ground, but to do so, there is a force being applied (gravity) over the given distance, therefore work is done.

I hope that answered the question!

 

[minutemen11]

you mean like when you have a capacitor??

if you have a positive charge and the capacitor is charged, the positive charge will go from low potential to high potential and the electrons will go from high potential to low potential. This is because a positive charge having low potential means its close to the negative plate, when current is run the force pushes it towards the (+) making it gain PE. This is the opposite in electrons..

Is this right? i hope i didnt just confuse myself.. good question tho!

 

[549]

if you have a positive charge and the capacitor is charged, the positive charge will go from low potential to high potential and the electrons will go from high potential to low potential. This is because a positive charge having low potential means its close to the negative plate, when current is run the force pushes it towards the (+) making it gain PE. This is the opposite in electrons..

I think it's the other way around. If you have a + charge, it will go from high V to low V. The opposite is true for an electron. The + charge will be accelerated to a negatively charged plate, therefore decreasing its PE and increasing its speed (and KE).

 

[thebillsfan]

alright, so does "positive potential"/"high potential" mean an area of high positive charge density? if that's the case, why would taking a negative charge from an area of low potential to high potential require work?

W = -∆PE = -q∆V

If you are going from an area of low potential to high potential, ∆V is +. q is -, as you stated. Therefore, W will be +.

shouldnt the negative charge willingy go where there is more positive charge?

It moves toward the area of high potential due to the electric force it feels, which is accelerating it. Therefore KE will be increasing, and that's where the work is done. Think about it this way maybe: if you drop a book from a height, it wants to go to the ground, but to do so, there is a force being applied (gravity) over the given distance, therefore work is done.

I hope that answered the question!

it does, but so it doesnt "take human work" to move an electron from high to low? if you have a positive charge at infinity and you move it to an area of very high potential, then you are giving it potential energy...yet negative work is being done to the positive charge?

 

[549]

Yeah, I guess now the question I have is what exactly does -W mean?! It seems it would be work done when a force is applied in the opposite direction of motion.

if you have a positive charge at infinity and you move it to an area of very high potential, then you are giving it potential energy...yet negative work is being done to the positive charge?

According the the equation relating W to ∆PE, increasing PE would cause a negative work to be done on a + charge. Is that because the force you'd have to apply is in the opposite direction it wants to go? That doesn't seem right at all.

Confusing.

 

[549]

Also, is there a difference in the way you look at the work depending on if work is transformed from KE to PE (as in the case of moving a + charge from high V to low V) or if you simply give it PE (moving said + charge from low V to high V), without any loss of KE?