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physics -- the potential energy of a proton released into an

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physics -- the potential energy of a proton released into an electric field DECREASES as it moves in the direction of the field. PE = -qEd. However, according to the high school teacher, the Voltage INCREASES as the proton moves into the field. V = -Ed. However, since BOTH Voltage and PE are in direct proportion to distance in these equations, it seems that both should decrease. Please explain in detail.
Submitted: 2 years ago.
Category: Homework
Expert:  David replied 2 years ago.

David :

the resolution is this. imagine a capacitor, left side is positively charged and right side is negative. Once the proton enters the field it will move to the right and accelerates so it indeed loses potential energy (as kinetic energy increases), note that in the first equation PE = -qEd, d is the distance between plates! and in the second part once proton enters the field it increases the electric field between itself and the right plate (as it adds mode positive charge). and note that V=-Ed is irrelevant here as this is NOT a voltage between the proton and the right plate, as d here is the distance from proton!

David :

These are tow completely different settings and meaning of the parameters are different, so there is no paradox at all :-)

David :

hope this helps

Customer:

what do you mean "adds more positive charge" -- don't understand how this "adds more positive charge"

David :

proton is positively charged, so it creates electric field in the direction of motion to the right of the proton and in the opposite direction to the left of the proton

David :

so the total electric field increases to the right of the proton and decreases to the left of the proton

David :

let me know if there are any questions, Ill be glad to clarify.

Customer:

well, I'm still stuck on the V = -Ed equation; if the total electric field increases as the proton moves toward the right, then shouldn't the Voltage decrease since V = negative Ed.

Customer:

or if we use V = (kq)/r and the distance increases, well, again, shouldn't the voltage decrease. So I still can't see from the EQUATIONS why the voltage should increase. In both cases, the equations lead one to belive that voltage decreases as the proton moves to the right. What am I missing

David :

Well let's look at this from another angle. V = -Ed hold only if the electric field is uniform!, for the proton it is not. so tis formula is simply not applicable to the proton.

Customer:

OK, that part makes perfect sense. But I'm still confused about V = (kq)/r ...... which is to say: Vr = kq ..... since V and r are inverse, does this mean that as the proton moves closer to the field (and the distance shrinks) the Voltage increases. Is that a correct interpretation?

David :

well, you need to be clear on the fact that V = (kq)/r tells you the potential at the point r meter from the proton!. it is positive!Not that d in the formula V=-Ed has nothing to do with this r. as d is the distance between the plates, not the distance from the proton.

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