# Two expressions for work - which is correct?

1. May 11, 2008

### Niles

When I want to find the energy in a system containing two points charges q and -q, I can use the following formula:

$$W = -Q\int_{\bf{a}}^{\bf{b}} {\bf{E}} \cdot d{\bf{l}} = Q(V(\mb b) - V(\mb a))$$

If my reference point is set to infinity, I can rewrite the last expression to:

$$W=QV(\mb r)$$

The two point charges are seperated by a distance of 2d, so the energy in the system calculated from the first formula will be (when bringing in the charge q in from infinity):

W = -K*q^2/(4d).

If I use the last expression, I get

W = -K*q^2/(2d).

Ehh, as far as I can see, I haven't made an obvious mistake. So how can this be?

2. May 11, 2008

### Hootenanny

Staff Emeritus
Why is this 4d of the charges are separated by a distance of 2d?

3. May 11, 2008

### Niles

The electric field at the point x=d is E = -K*q/(4d^2).

This I integrate, and it still gives me 4d.

4. May 11, 2008

### Hootenanny

Staff Emeritus
Let's step back a bit. Suppose our first charge is located at the origin, simply because it ismore conveinient. The the electric field due to this point charge is,

$$\mathbf{E} = k\frac{q}{R^2}\hat{\mathbf{r}}$$

Where R is the distance from the origin and $\hat{\mathbf{r}}$ is the radial unit vector. Agree?

Now let's bring in the second charge,

$$W = -Q\int^{2r}_{\infty}\mathbf{E}\cdot{d\mathbf{\ell}}$$

Now setting Q = -q and integrating along the (negative) radial direction,

$$W = qk\int^{2r}_{\infty}\frac{dR}{R^2}\left(-\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}\right)$$

$$W = -qk\int^{2r}_{\infty}\frac{dR}{R^2}$$

Do you agree?

5. May 11, 2008

### Niles

Yes, I do agree.

6. May 11, 2008

### Hootenanny

Staff Emeritus
Hence integrating,

$$W = qk\left[\frac{1}{R}\right]^{2r}_{\infty}$$

$$W = qk\left(\frac{1}{2r}-\lim_{R\to\infty}\frac{1}{R}\right)$$

$$W = \frac{kq}{2r}$$

7. May 11, 2008

### Niles

I agree with what you have done, but I also still agree on what I have done. What is wrong with my method?

8. May 11, 2008

### Hootenanny

Staff Emeritus

9. May 11, 2008

### Niles

I have my initial charge in x=-d and the other is brought in from infinity to x=d.

10. May 11, 2008

### Hootenanny

Staff Emeritus
Simply integrating this will not give you the potential energy at this point, you need to integrate before you substitute the limits in.

11. May 11, 2008

### Niles

Ok, this is how I've done:

We have that $$$W = - Q\int_\infty ^d {Edl = QV(d)}$$$

Approach 1, where $$$E = \frac{{ - kq}}{{\left( {2d'} \right)^2 }}$$$:

$$$W = - q\int_\infty ^d {\frac{{ - kq}}{{\left( {2d'} \right)^2 }}dd' = } \frac{{ - kq^2 }}{{4d}}$$$

Approach 2, where $$$V(d) = \frac{{ - kq}}{{2d}}$$$:

$$$W = q \cdot \frac{{ - kq}}{{2d}} = \frac{{ - kq^2 }}{{2d}}$$$

Can you point out the mistake here?

12. May 11, 2008

### Hootenanny

Staff Emeritus
Sorry Niles, I have no idea what your doing with you're first approach.

13. May 11, 2008

### Niles

Don't be sorry, sadly I do not explain things very well to others.

I found my error now. Instead of using an arbitrary distance in the electric field, I just inserted 2d, which is of course wrong, since I have to integrate it. This is of course also what you wrote in #10.

You did it again.