# Two expressions for work - which is correct?

## Main Question or Discussion Point

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?

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Hootenanny
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The two point charges are separated 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).
Why is this 4d of the charges are separated by a distance of 2d?

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

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

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

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

Yes, I do agree.

Hootenanny
Staff Emeritus
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Yes, I do agree.
Hence integrating,

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

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

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

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

Hootenanny
Staff Emeritus
Gold Member
I agree with what you have done, but I also still agree on what I have done. What is wrong with my method?

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

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

This I integrate, and it still gives me 4d.
Simply integrating this will not give you the potential energy at this point, you need to integrate before you substitute the limits in.

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?

Hootenanny
Staff Emeritus