Flux through a ring due to two charges

[palaphys]

Homework Statement

two charges +q and -q are placed on the axis of a ring at a distance 16cm and 9cm from the center on opposite sides. the ring has a charge 2q uniformly distributed on it. find the flux through the ring.

Relevant Equations

solid angle, gauss law

here is a simple diagram representing this situation. The issue I'm facing with this problem is a matter of sign convention.

We can write the flux through the ring as the sum of the contribution of fluxes due to the point charges individually.
So $$\Phi_{\text{net}} = \Phi_{+q} + \Phi_{-q} $$
additionally I was able to calculate the magnitudes of both of the individual fluxes as
$\left| \Phi_{+q} \right| = \frac{q}{10 \epsilon_0}$
and
$\left| \Phi_{-q} \right| = 2\frac{q}{10 \epsilon_0}$

both using a basic approach with solid angles(relating half angle to solid angle).but to get the net flux, should I add both the fluxes, or subtract them?

edit- radius of the ring is given as 12cm

 

[Steve4Physics]

Hi. A few thoughts…

Homework Statement: two charges +q and -q are placed on the axis of a ring at a distance 16cm and 9cm from the center on opposite sides. the ring has a charge 2q uniformly distributed on it. find the flux through the ring.

You haven’t given the radius of the ring.

We can write the flux through the ring as the sum of the contribution of fluxes due to the point charges individually.
So $$\Phi_{\text{net}} = \Phi_{+q} + \Phi_{-q} $$

You haven’t mentioned how you have dealt with the charge of 2q on the ring.

additionally I was able to calculate the magnitudes of both of the individual fluxes as
$\left| \Phi_{+q} \right| = \frac{q}{10 \epsilon_0}$
and
$\left| \Phi_{-q} \right| = 2\frac{q}{10 \epsilon_0}$

There's no way for us to check these without the radius of the ring.

to get the net flux, should I add both the fluxes, or subtract them?

What do you think? Hint: how would you find the individual and total fluxes through a surface enclosing charges of +q and -q?

 

[palaphys]

Hi. A few thoughts…


You haven’t given the radius of the ring.


You haven’t mentioned how you have dealt with the charge of 2q on the ring.


There's no way for us to check these without the radius of the ring.


What do you think? Hint: how would you find the individual and total fluxes through a surface enclosing charges of +q and -q?

sorry radius of ring is given as 12cm.
also for the flux calculation through the ring I ignored the charge on the ring, As I am unsure how to deal with self flux (if that exists) in electrostatics, and I am kind of confident that this question does not involve those complexities.(given its level)

As for the hint, I would think about the relative positioning of the E vector and A vector, and check for the sign of the dot product, but I am unable to do so.

 

[haruspex]

radius of ring is given as 12cm.

Then I confirm your results.

should I add both the fluxes, or subtract them?

What would the net flux be if they were both +q and equidistant on opposite sides?
(Consider symmetry.)

also for the flux calculation through the ring I ignored the charge on the ring,

Consider one point charge on the ring. What flux does it generate through the ring (and not in the plane of the ring)?

 

[Steve4Physics]

sorry radius of ring is given as 12cm.

In that case I agree with your values for the magnitudes of the fluxes. Though I would write '2/10' as '1/5'!

also for the flux calculation through the ring I ignored the charge on the ring, As I am unsure how to deal with self flux (if that exists) in electrostatics, and I am kind of confident that this question does not involve those complexities.(given its level)

Imagine lines of flux as lines leaving a positive charge or entering a negative charge (image from https://eee.poriyaan.in/topic/electric-flux-11760/):

If lines of flux pass through an area. that means they enter from one side of the area and leave from the other side of the area (image from https://artpictures.club/autumn-2023.html):

You need to imagine a ring made of lots of individual charges. Visualise the lines of flux from any single charge on the ring. Convince yourself that none of the lines of flux actually pass through area enclosed by the ring. So the flux from a ring of charge, through the area enclosed by the ring, is zero.

As for the hint, I would think about the relative positioning of the E vector and A vector, and check for the sign of the dot product, but I am unable to do so.

You are overcomplicating. Assuming you understand Gauss's law, try these questions:
Q1. A surface encloses a charge +q. What is the flux though the surface? (use Gauss's law).
Q2. A surface encloses a charge -q. What is the flux through the
surface? (use Gauss's law).
Q3. A surface encloses a charge +q and a charge -q:
a) What is the total charge enclosed?
b) What is the flux through the surface? (use Gauss's law).

Edit. Aha. @haruspex beat me too it.

 

[Gordianus]

We're dealing with an open surface and we should choose the normal vector (I know I'm a bit pedantic).

 

[palaphys]

What would the net flux be if they were both +q and equidistant on opposite sides?
(Consider symmetry.)

Intuition says zero.. flux due to one would "cancel out" the one due to the other

 

[palaphys]

We're dealing with an open surface and we should choose the normal vector (I know I'm a bit pedantic).

yes I wanted to do this

 

[palaphys]

You are overcomplicating. Assuming you understand Gauss's law, try these questions:
Q1. A surface encloses a charge +q. What is the flux though the surface? (use Gauss's law).
Q2. A surface encloses a charge -q. What is the flux through the
surface? (use Gauss's law).
Q3. A surface encloses a charge +q and a charge -q:
a) What is the total charge enclosed?
b) What is the flux through the surface? (use Gauss's law).

1.$\frac{q}{\epsilon_0}$
2. $\frac{-q}{\epsilon_0}$
3a)0 b)0

 

[Steve4Physics]

@palaphys, for an intuition, it helps to visualise flux lines. In the following diagram, the magnitude of the flux through the loop due to a charge +Q is ‘3 lines’.

For convenience, we can take ‘enter loop on left and leave loop on right’ as the positive sense here. This is equivalent to taking the normal of the loop’s area as pointing right. The value of the flux through the loop is +3 lines.

In the following diagram, the flux through the loop due to -Q is also ‘+3 lines’ because the lines of flux are also flowing left-to-right through the loop (in the same general direction as the normal).

With both charges, the total flux through the loop is (+3) + (+3) = +6 lines:

If you have flux lines flowing in opposite directions (e.g. with +Q and +Q symmetrically placed on opposite sides of the loop) you would have a flux of (+3) + (-3) = 0. Convince yourself of this if needed!

We may have to decide on the direction of the normal (i.e. the positive sense for flux). It depends on context:

For a closed surface (e.g. when using Gauss’s theorem) we take outwards as positive. So the normal of each surface element points outwards from the enclosed volume.

For open surfaces (e.g. the area enclosed by a loop) with a directed (oriented) area, we take the direction of the directed area’s normal as positive.

So, for a rigorous answer to your Post #1 question, you should say:
a) what direction you are using as the area's normal;
b) what values (not magnitudes) you have for $\Phi_{+q}$ and $\Phi_{-q}$ (based on a));
c) what the total is (based on a)).

 

[palaphys]

@palaphys, for an intuition, it helps to visualise flux lines. In the following diagram, the magnitude of the flux through the loop due to a charge +Q is ‘3 lines’.
View attachment 366657
For convenience, we can take ‘enter loop on left and leave loop on right’ as the positive sense here. This is equivalent to taking the normal of the loop’s area as pointing right. The value of the flux through the loop is +3 lines.

In the following diagram, the flux through the loop due to -Q is also ‘+3 lines’ because the lines of flux are also flowing left-to-right through the loop (in the same general direction as the normal).
View attachment 366659

With both charges, the total flux through the loop is (+3) + (+3) = +6 lines:
View attachment 366660
If you have flux lines flowing in opposite directions (e.g. with +Q and +Q symmetrically placed on opposite sides of the loop) you would have a flux of (+3) + (-3) = 0. Convince yourself of this if needed!

We may have to decide on the direction of the normal (i.e. the positive sense for flux). It depends on context:

For a closed surface (e.g. when using Gauss’s theorem) we take outwards as positive. So the normal of each surface element points outwards from the enclosed volume.

For open surfaces (e.g. the area enclosed by a loop) with a directed (oriented) area, we take the direction of the directed area’s normal as positive.

So, for a rigorous answer to your Post #1 question, you should say:
a) what direction you are using as the area's normal;
b) what values (not magnitudes) you have for $\Phi_{+q}$ and $\Phi_{-q}$ (based on a));
c) what the total is (based on a)).

This post makes everything clear. Very high quality. All my doubts on this question got cleared!
for a) I will take the direction of normal as rightward (as shown in diagram)
b) $\phi_{+q} = \frac{q}{10\varepsilon_0}$
and $\phi_{-q} = \frac{2q}{10\varepsilon_0}$
c) $\phi_{net}=\frac{3q}{10\varepsilon_0}$
(which matches the answer key given.)

Once again thanks a lot!