# Electric Potential Energy

1. Feb 28, 2016

### Jzhang27143

I am confused about the signs in calculating the potential energy change from the electrostatic force.

Suppose there was a point charge +q1 and I moved a second point charge +q2 from a distance of b from q1 to a distance of c from q1. c is smaller than b.

So the potential energy change is - ∫ F ⋅ ds where F is the electric force due to q1. F is radially outward and ds is dr radially inward (from b to c)\$, so F⋅ds = - F dr. So - ∫ F⋅ds = ∫ F dr = ∫ Kq1q2 / r^2 dr from b to c so the integral is Kq1q2 (-1/c + 1/b) which is negative. But the potential energy should be positive when q2 is brought closer to q1. Where is the sign error?

2. Feb 28, 2016

### drvrm

just check the integration part and substitution of values -i think there is a catch there!

3. Feb 28, 2016

### Jzhang27143

i'm not sure what the problem is with the integration. ∫ Kq1q2/r^2 dr from b to c is kq1 q2 ∫ dr/r^2 from b to c = kq1q2 (-1/r) from b to c = kq1q2(-1/c - (-1/b)) = kq1q2(-1/c + 1/b). Should i have done the integral from c to b? if so, why?

4. Feb 28, 2016

### Qwertywerty

You forget the negative sign outside the integral.
Also, calculating from point b to c, or vice versa, would depend on the question you'll be solving.

Hope this helps,
Qwertywerty.

5. Feb 28, 2016

### drvrm

well such a switch will correctly give you sign of P.E. to be + ve. but what does it mean?
i.e. how the P.E. is termed +ve or negative -
what is the concept of sign of P.E. related with work done by the field
or by an external agent carrying the task of doing work?

6. Feb 28, 2016

### Jzhang27143

well, I know that if an external agent is doing positive work, then the potential energy change is positive, and the potential energy change is negative if the external agent is doing negative work. In my example of moving q2 closer to q1, the work from the external agent should be positive and so should the potential energy change. However, I still don't understand why my calculations gave a negative change in potential energy.

7. Feb 28, 2016

### Qwertywerty

Last edited: Feb 28, 2016
8. Feb 28, 2016

### Qwertywerty

Edited.

9. Feb 28, 2016

### Jzhang27143

Oh, I thought I took care of the negative outside the integral though.

10. Feb 28, 2016

### Qwertywerty

'ds' represents displacement. So does 'dr'. Your assumption that F.ds = -F.dr, is wrong.

11. Feb 28, 2016

### Jzhang27143

Wait i thought the direction of ds was the direction of the path taken. Thats why I thought ds and the electric field pointed in opposite direction so the dot product was negative. If thats not the case, what determines the direction of ds?

12. Feb 28, 2016

### Qwertywerty

What you basically want to do is, calculate the negative of the work done by the field existing between the two charges.

When you calculate ∫F.dr, you are automatically taking into consideration the dot product while assigning the limits. It can be seen that, for example, WEF=F.Δr - Δr takes care of the sign, and the dot product simply serves to help provide a magnitude, and not a sign.

13. Feb 28, 2016

### Qwertywerty

'ds' does what you think it does. As in, ds represents the displacement in the direction you move. I have tried to clarify the dot product part in my previous reply.

14. Feb 28, 2016

### Jzhang27143

Oh, that makes sense now. So setting the limits from a larger to smaller distance by itself establishes that the electric force is in the opposite direction of the displacement, meaning that the negative from the dot product is redundant.

15. Feb 28, 2016

### Qwertywerty

Yes.

16. Feb 28, 2016

### drvrm

Your force should be negative gradient of potential so change in potential = (-) F. dr dr is not an increment but opposite to r a decrement in r suppose you are coming from infinity to b you should get a positive potential at b similarly at c and the difference of the two should be positive!
try to guess/find the error!