Finding Flux through a Disk due to Two Charges

[Saitama]

Homework Statement

Homework Equations

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!

 

[voko]

What is a field line?

 

[Saitama]

What is a field line?

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

 

[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?

 

[Saitama]

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?

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

Is this what you ask?

 

[voko]

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

Is this what you ask?

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 $\vec E$.

 

[Saitama]

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

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:
$$\vec{E}=\frac{kq}{|\vec{r}-\vec{p}|^3}(\vec{r}-\vec{p})-\frac{kq}{|\vec{r}-\vec{q}|^3}(\vec{r}-\vec{q})$$

Is this what you ask or did I still not answer your question? :)

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

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

 

[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 :)

 

[Saitama]

$$ {d \vec r \over ds} = {\vec E \over | \vec E | } $$

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

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

 

[voko]

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

 

[Saitama]

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

What do you mean?

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

 

[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.

 

[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.

 

[Saitama]

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.

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 $\phi_1$?

 

[TSny]

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 $\cos\theta=\frac{d}{\sqrt{D^2+d^2}}$.

Looks good to me.

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

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 $\alpha$?

 

[Saitama]

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 $\alpha$?

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

 

[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.

 

[Saitama]

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.

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. :)