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
In your method, where have you placed your initial charge?

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook