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What is the work done in shifting a charge of 2 Coulombs from pt A having potential +10V to pt B having a potential -5V? Answered

How will we calculate the potential difference? I'm not sure but I think the potential difference across AB is 5V.


The potential at A is higher than the potential at B

In fact, V(A) - V(B) = +10 - (-5) = 15 volts.

It will take work to move charge, positive charge, from B to A.

There is a mental picture I have for this.

To start with, I imagine there is a tiny basket of positive charges, sitting at point B, at the lower potential. There is also another similar tiny basket, sitting at point A.

Moreover, I imagine that I have some kind of magic rubber gloves, so I can pick the charges up, one by one, without charge flowing into my hands, and without changing the potentials that are there in the first place.

Yet the magic gloves will not help me to violate conservation of energy. So if I want to pick up a charge and move it, from one potential to another, then I might have to do some work on the charge, to make that happen.

Or I might have to accept work from the charge, if the change in potential is going the other way.

In the case of moving positive charges from B to A, from a low potential to a higher one, I would have do work on the charges. I would have to pull them loose from this negative potential at point B. Also I would have to push on them to move them closer to the positive potential at point A. How much work would this take? The answer is about 15 volts, or equivalently, 15 joules per coulomb.

The work needed to move 2 coulombs, from B to A, would be exactly 30 J.

If I want to move charges the other way, from A to B, the situation is a little different, because the charges naturally want to move that way.

In fact, those tiny imaginary baskets might have doors, or lids on them, for to keep the charges from flying loose in some circumstances.

And that's not a crazy idea. Consider the overhead luggage bins on airplanes. Those have doors on them, because the natural way the luggage wants to move is down, but the direction of down can change, relative to the walls of cabin, if the airplane decides to roll upside down, or sideways. The insides of airplane cabins also experience inertial pseudoforces, in addition to gravity. So a door on the overhead bin is good idea.

Anyway, getting back to moving charges, if I am going move charges from A to B, from a high potential to a lower one, then that is going to involve accepting work from the charges. Basically, I have to carefully unlatch the door on the basket at point A, and gently let out one positive charge, and accept energy from it, because this little charge is pushing against my hand the whole time, in its eagerness to get away from high potential, and to move towards low potential.

So in the process of moving charges from A to B, I have to accept work from the charges. Another way of saying this is, the charges do work on me, or equivalently, the work I do on the charges is negative.

So the amount of work I need to, on the charges, to move 2 coulombs of charge from A to B, is -30 J. Predictably, this is exactly -1 times the work needed to move the same amount of charge from B to A.

(Notice the minus sign. It is important.)

The potential difference is done by subtraction, and the signs are important. That should be enough of a clue for you to finish your homework.

Thanks..got the hint! It ain't mt homework though!

Yaya...I agree wid u!!

Wow great how people solve the homework problem of our students.
Why use schoolbooks or research if you have the great people at Instructables doing it all for you ;)

Haha..Its not my homework. I came across this question and was just wondering how to solve it.

Yes, sure, but I didn't ever cheat my tests.

Ah ha - I smell homework.