# Homework Help: Electricity contradiction? Which one is right

1. Apr 27, 2013

### x86

1. The problem statement, all variables and given/known data
The potential at a point is 20 V. Calculate the work done in bringing a charge of 0.5 C to this point.

2. Relevant equations
V = Ee / q
W = -Ee

3. The attempt at a solution
Ee = 10 J

W = -10 J

But the answer is 10 J. Why isn't it negative? It's contradicting what it taught me to do in previous questions.

The way I see it, work is being done on the system, therefore the work has to be negative (thats what it says in my book)

2. Apr 27, 2013

### barryj

Think about this. If you pushed a block on a flat surface with some friction, the work done would be the force X the distance and be positive. How is this different from when you bring a positive charge at infinity to a point where the voltage potential is +20 v, that is whether the work is positive or negative?

3. Apr 27, 2013

### x86

Well, it makes sense when you put it like that because now the way I see it is:

point 1 = infinity
point 2 = 20 V

----------------------------> force vector
So if you're going against the force vector, then cos180 = -1

So W = -Ee cos theta

W = +Ee

Is this correct? The answer is 10 J

So I guess the book is wrong, because you can't just say work is negative when force is being done against the system, since system is all about relativity

4. Apr 27, 2013

### barryj

One has to be careful about work being done by something or on something. When you compress a spring the force and direction are the same and you being the pusher does work on the spring. When you slowly release the compression the force is still toward the spring but the directioon is against the force and work is being done ON the pusher, yes?

5. Apr 27, 2013

### TSny

You have to be careful with these types of questions regarding the sign. Sometimes you might be asked for the work done by the electric force acting on the charge. At other times, you might be asked for the work done by you in moving the charge such that the charge does not accelerate. These works will have opposite sign. Of couse, you also must take into account the sign of the charge and the sign of the change in potential ΔV.

6. Apr 27, 2013

### barryj

Why would it be negative? Both charges are positive so they wouldn't they? If you started at infinity you would have to push the positive charge to the +20V potential wouldn't you? It would be like compressing a spring.

7. Apr 28, 2013

### haruspex

Can you post the section of the book that you consider has misled you?