# Electric potential energy

1. Mar 21, 2013

### assaftolko

1. The problem statement, all variables and given/known data

Two positive charges of q=3*10^-6 C are a distance 1 m apart.

What will be the work done by the electrical force in bringing one of the charges to a distance of 0.5 m from the other charge (which is fixed at its position through the whole process)?

2. Relevant equations

3. The attempt at a solution

Well from W=-ΔU I got w=-0.081 J which makes since because the electric force acts in contrast to the path of q as it moves closer to the other q - and so negative work is done. But when I tried to calculate the work directly from W1->2 = ∫$^{2}_{1}$Fel $\bullet$dr when 1=1m and 2=0.5m , I got the same result but positive - since the angle between Fel and dr is 180 deg... why is that? Isn't dr pointing to the direction of heading?

Last edited by a moderator: Mar 23, 2013
2. Mar 21, 2013

### rude man

Positive work is done ON the charge being moved.
The work done "by the electrical force" is the negative of the work done ON the charge.
In any case, the formula is W = +ΔU for the work done ON the charge.

The sign of ∫F dx will always be positive, where F is the force exerted ON the moved charge. This force is always the negative of the force exerted ON the moved charge BY the fixed charge. If the two charges are placed on the x axis, one at the origin and one at +1m, and we move the latter charge to the left, then the integral is ∫-F* -dx = +. If you move the charge at the origin instead the integral is ∫F* dx = + also. The integral is the work done ON the charge in either case.

The important point is to distinguish between the force exerted on the moved charge vs. the force the electric field exerts on the moved charge. One is the negative of the other, all the time.

3. Mar 21, 2013

### assaftolko

Im sorry but i dont see how the work is always positive when u calculate it through integration: in mechanics i got a lot of the times negative work through integration, in the case of friction for example and also for other general forces that we were asked to calculate their work

4. Mar 21, 2013

### rude man

If the force is + and the displacement is +, then ∫F*ds is +. This is the work done BY the force.

In the case of kinetic friction, the force is always COUNTER to the direction of motion so the work done BY the friction force is always negative. The work done by the force PUSHING THE OBJECT is positive.

It all depends on which force you're talking about to determine whether the integral is positive or negative.

5. Mar 21, 2013

### assaftolko

Ok but in this case the electrical force is outwards (repulsion) and the direction of motion is inwards (from 1m seperation to 0.5 seperation) so why the integration comes out positive??

6. Mar 21, 2013

### rude man

Which integration?

OK, you have q1 at x = 0 and q2 at x = 1m, and you are moving q2 from 1m to 0.5m.

If you integrate the force exerted by the field on the moved charge the integral is negative: ∫F*(-dx) < 0. If you're integrating the force needed to push the two charges together the integral is positive: ∫(-F)*(-dx) > 0.

7. Mar 21, 2013

### assaftolko

ok so i think that the problem is with my interpetion: they say in the question that at first they are 1m apart and then q2 is moved (let's say to the left towards q1) to 0.5m apart from q1. As I see it - the force of repulsion is positive with size kq1q2/r^2 and this is multiplied by dr and by cos180=-1: the cosine of the angle between F (repulsion) and dr (in the direction of movement, that is at the opposite direction of F). This integral is, as you pointed out, ∫-kq1q2/r^2 * dr - but if the integration limits are 1 for bottom and 0.5 for upper - you will get negative numerical result - you can check for yourself...

8. Mar 21, 2013

### rude man

Yes, you have chosen to compute the work done BY the electric field, which is why F is positive. And the work done by the field is negative as it must be. Otherwise the E field would produce positive work AND increase the potential energy of the system. That would violate the conservation of energy. So instead, change in potential energy + work done by the field = 0. ΔP.E. > 0, W < 0 where W = work done by the field.

The integral is positive only if F is negative and dr is negative, then this integral is the work done BY the force pushing q2 towards q1, threby increasing the P.E. of the system.

9. Mar 22, 2013

### assaftolko

No no no I'm sorry I got confused in my last post - I meant to say in the last sentence that you will get positive numerical result, and not negetive like you should:

∫-kq1q2/r^2 * dr = -kq1q2∫dr/r^2 = -kq1q2*-1/r = -kq1q2*(-1/0.5-(-1/1))=+0.081 J

10. Mar 22, 2013

### assaftolko

It comes out positive: $\vec{F}$$\bullet$$\vec{dr}$ = Fdrcos180 = -Fdr. If you integrate this expression between r1=1m and r2=0.5m you will get positive numerical result!

F is the coloumb force, and this force acts to the right. dr is a displacment element in the direction of the displacment - so it's to the left: The size of both vectors is positive, and the angle between them at all times is 180 deg

Last edited: Mar 22, 2013
11. Mar 22, 2013

### rude man

OK, let's say you're moving q1 instead of q2. q1 is at x = -1 and q2 is at x = 0. Then dx > 0 for sure.
F on q1 due to q2 = -kq1*q2/x2. So ∫Fdx = kq1*q2/x which evaluated from -1 to -0.5 is kq1*q2(1/-0.5 - 1/-1) = kq1*q2[-2 - (-1)] = -kq1*q2 < 0.

This is no different than putting q1 at 0 and q2 at x = 1, then moving q2 to the left:
F on q2 due to q1 = kq1*q2/x2, ∫Fdx = -kq1*q2/x which evaluated from 1 to 0.5 is -kq1*q2(1/0.5 - 1/1) = -kq1*q2 < 0 also. Notice that the limits of integration go + in the first case and - in the second case. That takes care of the direction of motion being + in the 1st case and - in the second.

12. Mar 22, 2013

### assaftolko

My friend it seems to me that you integrated 1/x^2 as 1/x and not -1/x...

13. Mar 22, 2013

### rude man

Where?

14. Mar 22, 2013

### assaftolko

In your last paragraph: you wrote ...(1/0.5 - 1/1), but it supoose to be (-1/0.5-(-1/1))

15. Mar 22, 2013

### rude man

No. Read the last paragraph again. q1 is at x=0 and q2 is at x = +1. q2 is moved from x=1 to x=0.5. So the limits are 1 and 0.5.

16. Mar 23, 2013

### assaftolko

I agree on the limits, but i dont get why you dont have a minus sign before 1/0.5 for example, since you should put the limit 0.5 into the expression -1/x and not 1/x...

bottom limit: 1m, upper limit: 0.5m. 1/x^2 integrates as -1/x. When you integrate you calculate the upper limit of integration before subtracting from it the bottom limit of integration, so you get: -1/0.5-(-1/1)= -1. This -1 is multiplied by the -1 that comes from cos180, and so we get 1. The other numerical values are of q1, q2 and k and are all positive also - so at the end we get a positive number:
$\int^{0.5}_{1}$$\vec{F}$$\bullet$$\vec{dx}$ = $\int$$^{0.5}_{1}$Fdxcos180 = $\int$$^{0.5}_{1}$kq1q2/x^2*dx*-1 = -kq1q2$\int^{0.5}_{1}$1/x^2*dx = -kq1q2*(-1/x)$\left|$$^{0.5}_{1}$ = -kq1q2*(-1/0.5--1/1) = -kq1q2*-1 = kq1q2 = +

i dont get why this isnt true...

Last edited: Mar 23, 2013
17. Mar 23, 2013

### rude man

You are contradicting yourself. You're putting in negative limits when they are clearly positive.

You get 1/.5 - 1/1= 1, not -1/.5 - 1/1 = -1.

18. Mar 23, 2013

### assaftolko

No i'm not! I'm putting positive limits, the minus comes from -1/x!

19. Mar 23, 2013

### vela

Staff Emeritus
You just have to be a little more careful with the signs. You have
$$dW = \vec{F}\cdot d\vec{r} = \|\vec{F}\| \|d\vec{r}\| \cos(180^\circ) = -\|\vec{F}\| \|d\vec{r}\|.$$ Clearly, that quantity is negative as you'd expect. The problem arises when you change $\|d\vec{r}\|$ to $dr$. Because the lower limit is greater than the upper limit, dr<0, so you have to have $\|d\vec{r}\| = -dr$.

20. Mar 23, 2013

### assaftolko

What? But why? Isnt dr pointing in the direction of the displacement? I thought that when i write the expression with cos180 then the vector dr just becomes to be dr which is its size - a positive number