# Spacing of field lines

1. Feb 22, 2014

### Saitama

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

2. Relevant equations

3. The attempt at a solution
The given field lines are at a maximum distance exactly midway between the charges but I don't know which equation to use to find it.

Any help is appreciated. Thanks!

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2. Feb 22, 2014

### voko

What is a field line?

3. Feb 22, 2014

### Saitama

It represents the direction of force acting on the charged particle in an electric field.

4. Feb 22, 2014

### voko

Yes, but how is it defined, exactly?

Let the +q charge be the origin. At location $\vec r$, what is the direction of the force? What property of the line is related with this direction?

5. Feb 22, 2014

### Saitama

$$\vec{E}=\frac{\vec{F}}{q}$$
(where F is the force on charge)

6. Feb 22, 2014

### voko

No. In fact, you did not answer any of my questions :)

You did not show what the direction of $\vec E$ really is at the given $\vec r$ in this problem.

Nor did you explain what property of the line is related to the direction of $\vec E$.

7. Feb 22, 2014

### Saitama

Do you want me to evaluate an expression for electric field at the given $\vec{r}$? If so, this is my attempt:

Let the positive charge be at $\vec{p}$ and negative charge at $\vec{q}$. The net field at $\vec{r}$ is given by:
$$\vec{E}=\frac{kq}{|\vec{r}-\vec{p}|^3}(\vec{r}-\vec{p})-\frac{kq}{|\vec{r}-\vec{q}|^3}(\vec{r}-\vec{q})$$

$\vec{E}$ is tangent to field line.

8. Feb 22, 2014

### voko

So, if we parametrise filed lines with natural parameter $s$, we should have $${d \vec r \over ds} = {\vec E \over | \vec E | }$$ Solve :)

9. Feb 22, 2014

### Saitama

Erm....is there no other way to solve the problem?

I feel $|\vec{E}|$ would be very messy.

10. Feb 22, 2014

### voko

You can get rid of $|\vec E|$ by choosing two components of $\vec r$, say x and y coordinates, then $${ dy \over dx } = { {dy \over ds} \over {dx \over ds}} = ...$$

11. Feb 22, 2014

### Saitama

What do you mean? :uhh:

Do you ask me to write $\vec{r}=x\hat{i}+y\hat{j}$?

12. Feb 22, 2014

### voko

Well, I did not ask. I suggested that you could :) You can decompose the position vector in any two components and still get rid of the magnitude of the field.

13. Feb 22, 2014

### TSny

Maybe it would help to think about the electric flux through a circular area of diameter D centered on the axis of the dipole midway between the charges and perpendicular to the axis.

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14. Feb 22, 2014

### Saitama

Ah, nice. :)

The flux leaving from +q making a cone of angle of $90^{\circ}$ is given by
$$\phi_1=\frac{q}{2\epsilon_o}\left(1-\frac{1}{\sqrt{2}}\right)$$
The flux through the circle of diameter D is:
$$\phi_2=2\times \frac{q}{2\epsilon_0}\left(1-\cos\theta\right)$$
where $\cos\theta=\frac{d}{\sqrt{D^2+d^2}}$.

I can equate them to obtain D but I am not sure about $\phi_1$. Is there no contribution from -q in $\phi_1$?

Last edited: Feb 22, 2014
15. Feb 23, 2014

### TSny

Looks good to me.

The flux through the disk can be thought of as the total number of field lines piercing the disk. Are there any lines that pierce the disk other than the lines that leave the +q charge within the cone of half-angle $\alpha$?

16. Feb 23, 2014

### Saitama

$\phi_2$ is through disk you have shown in the sketch. Both the charges contribute to $\phi_2$, am I right in saying this?

17. Feb 23, 2014

### TSny

Yes, both charges contribute to the electric field at the disk. $\phi_2$ is the flux through the disk due to the net E field at the disk and therefore represents the flux through the disk due to both charges. I believe your expression for $\phi_2$ is correct.

For $\phi_1$ you are following the curved field lines. These correspond to the net field of both charges. (But very close to $q_1$ the net field is essentially the same as the field of $q_1$ alone.) The number of curved field lines that pierce the disk is the same as the number of field lines that come off of $q_1$ within the cone. I believe your expression for $\phi_1$ is also correct.

18. Feb 23, 2014

### Saitama

Thanks TSny! I have reached the correct answer 30.52 cm.

That was a very neat way to solve the problem, I will be making a note of this. :)